For radially freely falling bodies, it holds that θ ˙ = 0 θ ˙ = 0 theta^(˙)=0\dot{\theta}=0θ˙=0 and ϕ ˙ = 0 ϕ ˙ = 0 phi^(˙)=0\dot{\phi}=0ϕ˙=0, which means that the Lagrangian reduces to
(3.1370) L = ( 1 r r ) ( x ˙ 0 ) 2 ( 1 r r ) 1 r ˙ 2 . (3.1370) L = 1 r r x ˙ 0 2 1 r r 1 r ˙ 2 . {:(3.1370)L=(1-(r_(**))/(r))(x^(˙)^(0))^(2)-(1-(r_(**))/(r))^(-1)r^(˙)^(2).:}\begin{equation*} \mathcal{L}=\left(1-\frac{r_{*}}{r}\right)\left(\dot{x}^{0}\right)^{2}-\left(1-\frac{r_{*}}{r}\right)^{-1} \dot{r}^{2} . \tag{3.1370} \end{equation*}(3.1370)L=(1rr)(x˙0)2(1rr)1r˙2.
Now, the geodesic equations for x 0 x 0 x^(0)x^{0}x0 and r r rrr coordinates are
(3.1371) L x 0 d d τ L x ˙ 0 = 2 d d τ [ ( 1 r r ) x ˙ 0 ] = 0 (3.1372) L r d d τ L r ˙ = r r 2 ( x ˙ 0 ) 2 ( 1 r r ) 2 r r 2 r ˙ 2 + 2 d d τ [ ( 1 r r ) 1 r ˙ ] = 0 (3.1371) L x 0 d d τ L x ˙ 0 = 2 d d τ 1 r r x ˙ 0 = 0 (3.1372) L r d d τ L r ˙ = r r 2 x ˙ 0 2 1 r r 2 r r 2 r ˙ 2 + 2 d d τ 1 r r 1 r ˙ = 0 {:[(3.1371)(delL)/(delx^(0))-(d)/(d tau)(delL)/(delx^(˙)^(0))=-2(d)/(d tau)[(1-(r_(**))/(r))x^(˙)^(0)]=0],[(3.1372)(delL)/(del r)-(d)/(d tau)(delL)/(del(r^(˙)))=(r_(**))/(r^(2))(x^(˙)^(0))^(2)-(1-(r_(**))/(r))^(-2)(r_(**))/(r^(2))r^(˙)^(2)+2(d)/(d tau)[(1-(r_(**))/(r))^(-1)(r^(˙))]=0]:}\begin{align*} \frac{\partial \mathcal{L}}{\partial x^{0}}-\frac{d}{d \tau} \frac{\partial \mathcal{L}}{\partial \dot{x}^{0}} & =-2 \frac{d}{d \tau}\left[\left(1-\frac{r_{*}}{r}\right) \dot{x}^{0}\right]=0 \tag{3.1371}\\ \frac{\partial \mathcal{L}}{\partial r}-\frac{d}{d \tau} \frac{\partial \mathcal{L}}{\partial \dot{r}} & =\frac{r_{*}}{r^{2}}\left(\dot{x}^{0}\right)^{2}-\left(1-\frac{r_{*}}{r}\right)^{-2} \frac{r_{*}}{r^{2}} \dot{r}^{2}+2 \frac{d}{d \tau}\left[\left(1-\frac{r_{*}}{r}\right)^{-1} \dot{r}\right]=0 \tag{3.1372} \end{align*}(3.1371)Lx0ddτLx˙0=2ddτ[(1rr)x˙0]=0(3.1372)LrddτLr˙=rr2(x˙0)2(1rr)2rr2r˙2+2ddτ[(1rr)1r˙]=0
The geodesic equation for the x 0 x 0 x^(0)x^{0}x0-coordinate leads to
(3.1373) ( 1 r r ) x ˙ 0 = E = const. (3.1373) 1 r r x ˙ 0 = E =  const.  {:(3.1373)(1-(r_(**))/(r))x^(˙)^(0)=E=" const. ":}\begin{equation*} \left(1-\frac{r_{*}}{r}\right) \dot{x}^{0}=E=\text { const. } \tag{3.1373} \end{equation*}(3.1373)(1rr)x˙0=E= const. 
In addition, we can find more information about the geodesics by using that
(3.1374) d L d τ = 0 (3.1374) d L d τ = 0 {:(3.1374)(dL)/(d tau)=0:}\begin{equation*} \frac{d \mathcal{L}}{d \tau}=0 \tag{3.1374} \end{equation*}(3.1374)dLdτ=0
Thus, we have that
(3.1375) L = ( 1 r r ) ( x ˙ 0 ) 2 ( 1 r r ) 1 r ˙ 2 = ϵ = const. (3.1375) L = 1 r r x ˙ 0 2 1 r r 1 r ˙ 2 = ϵ =  const.  {:(3.1375)L=(1-(r_(**))/(r))(x^(˙)^(0))^(2)-(1-(r_(**))/(r))^(-1)r^(˙)^(2)=-epsilon=" const. ":}\begin{equation*} \mathcal{L}=\left(1-\frac{r_{*}}{r}\right)\left(\dot{x}^{0}\right)^{2}-\left(1-\frac{r_{*}}{r}\right)^{-1} \dot{r}^{2}=-\epsilon=\text { const. } \tag{3.1375} \end{equation*}(3.1375)L=(1rr)(x˙0)2(1rr)1r˙2=ϵ= const. 
Combining the two expressions defining the quantities E E EEE and ϵ ϵ epsilon\epsilonϵ, we find that
(3.1376) r ˙ 2 = ϵ ( 1 r r ) + E 2 . (3.1376) r ˙ 2 = ϵ 1 r r + E 2 . {:(3.1376)r^(˙)^(2)=epsilon(1-(r_(**))/(r))+E^(2).:}\begin{equation*} \dot{r}^{2}=\epsilon\left(1-\frac{r_{*}}{r}\right)+E^{2} . \tag{3.1376} \end{equation*}(3.1376)r˙2=ϵ(1rr)+E2.
In order to find r ( τ ) r ( τ ) r(tau)r(\tau)r(τ), we solve for r ˙ r ˙ r^(˙)\dot{r}r˙, separate the variables τ τ tau\tauτ and r r rrr, and integrate both variables, i.e.,
d r d τ = ϵ ( 1 r r ) + E 2 d τ = d r ϵ ( 1 r r ) + E 2 (3.1377) τ = 0 τ d τ = d r ϵ ( 1 r r ) + E 2 d r d τ = ϵ 1 r r + E 2 d τ = d r ϵ 1 r r + E 2 (3.1377) τ = 0 τ d τ = d r ϵ 1 r r + E 2 {:[(dr)/(d tau)=-sqrt(epsilon(1-(r_(**))/(r))+E^(2))=>d tau=-(dr)/(sqrt(epsilon(1-(r_(**))/(r))+E^(2)))],[(3.1377)=>tau=int_(0)^(tau)dtau^(')=-int(dr)/(sqrt(epsilon(1-(r_(**))/(r))+E^(2)))]:}\begin{align*} \frac{d r}{d \tau}=-\sqrt{\epsilon\left(1-\frac{r_{*}}{r}\right)+E^{2}} & \Rightarrow d \tau=-\frac{d r}{\sqrt{\epsilon\left(1-\frac{r_{*}}{r}\right)+E^{2}}} \\ & \Rightarrow \tau=\int_{0}^{\tau} d \tau^{\prime}=-\int \frac{d r}{\sqrt{\epsilon\left(1-\frac{r_{*}}{r}\right)+E^{2}}} \tag{3.1377} \end{align*}drdτ=ϵ(1rr)+E2dτ=drϵ(1rr)+E2(3.1377)τ=0τdτ=drϵ(1rr)+E2
which describes the spacetime curves. Note that we use a minus sign for the square root, since the bodies are radially freely falling toward r = 0 r = 0 r=0r=0r=0, i.e., r ˙ < 0 r ˙ < 0 r^(˙) < 0\dot{r}<0r˙<0.
Usually, when solving equations of this type, one tries to find a solution r = r ( τ ) r = r ( τ ) r=r(tau)r=r(\tau)r=r(τ), but one can also find a solution τ = τ ( r ) τ = τ ( r ) tau=tau(r)\tau=\tau(r)τ=τ(r), since r ( τ ) r ( τ ) r(tau)r(\tau)r(τ) is one-to-one. In principle, one can then find x 0 x 0 x^(0)x^{0}x0 as
(3.1378) x 0 ( τ ) = 0 τ [ 1 r r ( τ ) ] 1 E d τ (3.1378) x 0 ( τ ) = 0 τ 1 r r τ 1 E d τ {:(3.1378)x^(0)(tau)=int_(0)^(tau)[1-(r_(**))/(r(tau^(')))]^(-1)Edtau^('):}\begin{equation*} x^{0}(\tau)=\int_{0}^{\tau}\left[1-\frac{r_{*}}{r\left(\tau^{\prime}\right)}\right]^{-1} E d \tau^{\prime} \tag{3.1378} \end{equation*}(3.1378)x0(τ)=0τ[1rr(τ)]1Edτ
By using the Euler-Lagrange equations in the same manner as in Problem 2.70, the equations of motion are given by
(3.1379) d d s ( 1 r r ) t ˙ = 0 (3.1380) d d s r 2 ϕ ˙ = 0 ( 1 r r ) t ˙ = E (3.1381) r r 2 t ˙ 2 2 r ϕ ˙ 2 = r r 2 ( 1 r r ) 2 r ˙ 2 + 2 d d s [ ( 1 r r ) 1 r ˙ ] (3.1379) d d s 1 r r t ˙ = 0 (3.1380) d d s r 2 ϕ ˙ = 0 1 r r t ˙ = E (3.1381) r r 2 t ˙ 2 2 r ϕ ˙ 2 = r r 2 1 r r 2 r ˙ 2 + 2 d d s 1 r r 1 r ˙ {:[(3.1379)(d)/(ds)(1-(r_(**))/(r))t^(˙)=0],[(3.1380)(d)/(ds)r^(2)phi^(˙)=0quad=>quad(1-(r_(**))/(r))t^(˙)=E],[(3.1381)(r_(**))/(r^(2))*t^(˙)^(2)-2rphi^(˙)^(2)=(r_(**))/(r^(2))(1-(r_(**))/(r))^(-2)r^(˙)^(2)+2(d)/(ds)[(1-(r_(**))/(r))^(-1)(r^(˙))]]:}\begin{align*} \frac{d}{d s}\left(1-\frac{r_{*}}{r}\right) \dot{t} & =0 \tag{3.1379}\\ \frac{d}{d s} r^{2} \dot{\phi} & =0 \quad \Rightarrow \quad\left(1-\frac{r_{*}}{r}\right) \dot{t}=E \tag{3.1380}\\ \frac{r_{*}}{r^{2}} \cdot \dot{t}^{2}-2 r \dot{\phi}^{2} & =\frac{r_{*}}{r^{2}}\left(1-\frac{r_{*}}{r}\right)^{-2} \dot{r}^{2}+2 \frac{d}{d s}\left[\left(1-\frac{r_{*}}{r}\right)^{-1} \dot{r}\right] \tag{3.1381} \end{align*}(3.1379)dds(1rr)t˙=0(3.1380)ddsr2ϕ˙=0(1rr)t˙=E(3.1381)rr2t˙22rϕ˙2=rr2(1rr)2r˙2+2dds[(1rr)1r˙]
where E E EEE and h h hhh are constants.
For circular orbits, r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0 is constant. From the third of the equations of motion, we then obtain
(3.1382) r r 0 2 i 2 = 2 r 0 ϕ ˙ 2 Δ t = Δ ϕ 2 r 0 3 r (3.1382) r r 0 2 i 2 = 2 r 0 ϕ ˙ 2 Δ t = Δ ϕ 2 r 0 3 r {:(3.1382)(r_(**))/(r_(0)^(2))i^(2)=2r_(0)phi^(˙)^(2)quad=>quad Delta t=Delta phisqrt((2r_(0)^(3))/(r_(**))):}\begin{equation*} \frac{r_{*}}{r_{0}^{2}} i^{2}=2 r_{0} \dot{\phi}^{2} \quad \Rightarrow \quad \Delta t=\Delta \phi \sqrt{\frac{2 r_{0}^{3}}{r_{*}}} \tag{3.1382} \end{equation*}(3.1382)rr02i2=2r0ϕ˙2Δt=Δϕ2r03r
By inserting this into the line element d s 2 d s 2 ds^(2)d s^{2}ds2, we obtain
(3.1383) Δ s 2 = 2 r 0 2 ( r 0 r 3 2 ) Δ ϕ 2 (3.1383) Δ s 2 = 2 r 0 2 r 0 r 3 2 Δ ϕ 2 {:(3.1383)Deltas^(2)=2r_(0)^(2)((r_(0))/(r_(**))-(3)/(2))Deltaphi^(2):}\begin{equation*} \Delta s^{2}=2 r_{0}^{2}\left(\frac{r_{0}}{r_{*}}-\frac{3}{2}\right) \Delta \phi^{2} \tag{3.1383} \end{equation*}(3.1383)Δs2=2r02(r0r32)Δϕ2
Since Δ s 2 0 Δ s 2 0 Deltas^(2) >= 0\Delta s^{2} \geq 0Δs20 for any physically viable worldline, we obtain r 0 3 r / 2 r 0 3 r / 2 r_(0) >= 3r_(**)//2r_{0} \geq 3 r_{*} / 2r03r/2. Thus, there are no circular orbits inside the r = 3 r / 2 r = 3 r / 2 r=3r_(**)//2r=3 r_{*} / 2r=3r/2.

2.75

Geodesics for massive test objects in the Schwarzschild metric satisfy the equation of motion
(3.1384) E 2 r ˙ 2 2 = 1 2 ( 1 + L 2 r 2 ) ( 1 r r ) = V ( r ) (3.1384) E 2 r ˙ 2 2 = 1 2 1 + L 2 r 2 1 r r = V ( r ) {:(3.1384)E^(2)-(r^(˙)^(2))/(2)=(1)/(2)(1+(L^(2))/(r^(2)))(1-(r_(**))/(r))=V(r):}\begin{equation*} E^{2}-\frac{\dot{r}^{2}}{2}=\frac{1}{2}\left(1+\frac{L^{2}}{r^{2}}\right)\left(1-\frac{r_{*}}{r}\right)=V(r) \tag{3.1384} \end{equation*}(3.1384)E2r˙22=12(1+L2r2)(1rr)=V(r)
where L = r 2 φ ˙ L = r 2 φ ˙ L=r^(2)varphi^(˙)L=r^{2} \dot{\varphi}L=r2φ˙ is the angular momentum. Approximately circular orbits have an energy close to the minimum of V ( r ) V ( r ) V(r)V(r)V(r). To find this minimum, we define U ( x ) = U ( x ) = U(x)=U(x)=U(x)= V ( 1 x ) V 1 x V((1)/(x))V\left(\frac{1}{x}\right)V(1x) and set U ( x 0 ) = 0 set U x 0 = 0 setU^(')(x_(0))=0\operatorname{set} U^{\prime}\left(x_{0}\right)=0setU(x0)=0 :
(3.1385) U ( x 0 ) = x 0 L 2 ( 1 r x 0 ) r 2 ( 1 + L 2 x 0 2 ) = 0 (3.1385) U x 0 = x 0 L 2 1 r x 0 r 2 1 + L 2 x 0 2 = 0 {:(3.1385)U^(')(x_(0))=x_(0)L^(2)(1-r_(**)x_(0))-(r_(**))/(2)(1+L^(2)x_(0)^(2))=0:}\begin{equation*} U^{\prime}\left(x_{0}\right)=x_{0} L^{2}\left(1-r_{*} x_{0}\right)-\frac{r_{*}}{2}\left(1+L^{2} x_{0}^{2}\right)=0 \tag{3.1385} \end{equation*}(3.1385)U(x0)=x0L2(1rx0)r2(1+L2x02)=0
with the solution
(3.1386) x 0 = 1 3 r ( 1 1 3 r 2 L 2 ) (3.1386) x 0 = 1 3 r 1 1 3 r 2 L 2 {:(3.1386)x_(0)=(1)/(3r_(**))(1-sqrt(1-(3r_(**)^(2))/(L^(2)))):}\begin{equation*} x_{0}=\frac{1}{3 r_{*}}\left(1-\sqrt{1-\frac{3 r_{*}^{2}}{L^{2}}}\right) \tag{3.1386} \end{equation*}(3.1386)x0=13r(113r2L2)
where the solution with a plus sign in front of the square root is unstable. For r 0 = 1 x 0 r r 0 = 1 x 0 r r_(0)=(1)/(x_(0))≫r_(**)r_{0}=\frac{1}{x_{0}} \gg r_{*}r0=1x0r, we require that L 2 r 2 L 2 r 2 L^(2)≫r_(**)^(2)L^{2} \gg r_{*}^{2}L2r2. The second derivative of the potential is
(3.1387) V ( r ) = d 2 d r 2 U ( x ) = d d r ( d x d r U ( x ) ) = ( d x d r ) 2 U ( x ) + d 2 x d r 2 U ( x ) (3.1387) V ( r ) = d 2 d r 2 U ( x ) = d d r d x d r U ( x ) = d x d r 2 U ( x ) + d 2 x d r 2 U ( x ) {:(3.1387)V^('')(r)=(d^(2))/(dr^(2))U(x)=(d)/(dr)*((dx)/(dr)U^(')(x))=((dx)/(dr))^(2)U^('')(x)+(d^(2)x)/(dr^(2))U^(')(x):}\begin{equation*} V^{\prime \prime}(r)=\frac{d^{2}}{d r^{2}} U(x)=\frac{d}{d r} \cdot\left(\frac{d x}{d r} U^{\prime}(x)\right)=\left(\frac{d x}{d r}\right)^{2} U^{\prime \prime}(x)+\frac{d^{2} x}{d r^{2}} U^{\prime}(x) \tag{3.1387} \end{equation*}(3.1387)V(r)=d2dr2U(x)=ddr(dxdrU(x))=(dxdr)2U(x)+d2xdr2U(x)
At r 0 = 1 x 0 r 0 = 1 x 0 r_(0)=(1)/(x_(0))r_{0}=\frac{1}{x_{0}}r0=1x0, we therefore obtain
(3.1388) V ( r 0 ) = ( d x d r ) 2 U ( x ) + 0 = x 0 4 U ( x 0 ) = U ( x 0 ) r 0 4 (3.1388) V r 0 = d x d r 2 U ( x ) + 0 = x 0 4 U x 0 = U x 0 r 0 4 {:(3.1388)V^('')(r_(0))=((dx)/(dr))^(2)U^('')(x)+0=x_(0)^(4)U^('')(x_(0))=(U^('')(x_(0)))/(r_(0)^(4)):}\begin{equation*} V^{\prime \prime}\left(r_{0}\right)=\left(\frac{d x}{d r}\right)^{2} U^{\prime \prime}(x)+0=x_{0}^{4} U^{\prime \prime}\left(x_{0}\right)=\frac{U^{\prime \prime}\left(x_{0}\right)}{r_{0}^{4}} \tag{3.1388} \end{equation*}(3.1388)V(r0)=(dxdr)2U(x)+0=x04U(x0)=U(x0)r04
We find that
(3.1389) U ( x 0 ) = 3 r L 2 x 0 + L 2 = L 2 L 2 ( 1 1 3 r 2 L 2 ) = L 2 1 3 r 2 L 2 (3.1389) U x 0 = 3 r L 2 x 0 + L 2 = L 2 L 2 1 1 3 r 2 L 2 = L 2 1 3 r 2 L 2 {:(3.1389)U^('')(x_(0))=-3r_(**)L^(2)x_(0)+L^(2)=L^(2)-L^(2)(1-sqrt(1-(3r_(**)^(2))/(L^(2))))=L^(2)sqrt(1-(3r_(**)^(2))/(L^(2))):}\begin{equation*} U^{\prime \prime}\left(x_{0}\right)=-3 r_{*} L^{2} x_{0}+L^{2}=L^{2}-L^{2}\left(1-\sqrt{1-\frac{3 r_{*}^{2}}{L^{2}}}\right)=L^{2} \sqrt{1-\frac{3 r_{*}^{2}}{L^{2}}} \tag{3.1389} \end{equation*}(3.1389)U(x0)=3rL2x0+L2=L2L2(113r2L2)=L213r2L2
Thus, for small ρ = r r 0 ρ = r r 0 rho=r-r_(0)\rho=r-r_{0}ρ=rr0, we find that
(3.1390) V ( r ) V 0 + 1 2 U ( x 0 ) r 0 4 ρ 2 (3.1390) V ( r ) V 0 + 1 2 U x 0 r 0 4 ρ 2 {:(3.1390)V(r)≃V_(0)+(1)/(2)(U^('')(x_(0)))/(r_(0)^(4))*rho^(2):}\begin{equation*} V(r) \simeq V_{0}+\frac{1}{2} \frac{U^{\prime \prime}\left(x_{0}\right)}{r_{0}^{4}} \cdot \rho^{2} \tag{3.1390} \end{equation*}(3.1390)V(r)V0+12U(x0)r04ρ2
The equation of motion for ρ ρ rho\rhoρ becomes
(3.1391) ρ ¨ + U ( x 0 ) r 0 4 ρ = ρ ¨ + ω 2 ρ = 0 (3.1391) ρ ¨ + U x 0 r 0 4 ρ = ρ ¨ + ω 2 ρ = 0 {:(3.1391)rho^(¨)+(U^('')(x_(0)))/(r_(0)^(4))rho=rho^(¨)+omega^(2)rho=0:}\begin{equation*} \ddot{\rho}+\frac{U^{\prime \prime}\left(x_{0}\right)}{r_{0}^{4}} \rho=\ddot{\rho}+\omega^{2} \rho=0 \tag{3.1391} \end{equation*}(3.1391)ρ¨+U(x0)r04ρ=ρ¨+ω2ρ=0
The proper time period for small radial oscillations is therefore
(3.1392) T ρ = 2 π ω = 2 π r 0 2 U ( x 0 ) = 2 π x 0 2 U ( x 0 ) = 9 π r 2 2 L [ 1 + 9 r 2 4 L 2 + O ( r L ) 4 ] (3.1392) T ρ = 2 π ω = 2 π r 0 2 U x 0 = 2 π x 0 2 U x 0 = 9 π r 2 2 L 1 + 9 r 2 4 L 2 + O r L 4 {:(3.1392)T_(rho)=(2pi)/(omega)=(2pir_(0)^(2))/(sqrt(U^('')(x_(0))))=(2pi)/(x_(0)^(2)sqrt(U^('')(x_(0))))=(9pir_(**)^(2))/(2L)[1+(9r_(**)^(2))/(4L^(2))+O((r_(**))/(L))^(4)]:}\begin{equation*} T_{\rho}=\frac{2 \pi}{\omega}=\frac{2 \pi r_{0}^{2}}{\sqrt{U^{\prime \prime}\left(x_{0}\right)}}=\frac{2 \pi}{x_{0}^{2} \sqrt{U^{\prime \prime}\left(x_{0}\right)}}=\frac{9 \pi r_{*}^{2}}{2 L}\left[1+\frac{9 r_{*}^{2}}{4 L^{2}}+\mathcal{O}\left(\frac{r_{*}}{L}\right)^{4}\right] \tag{3.1392} \end{equation*}(3.1392)Tρ=2πω=2πr02U(x0)=2πx02U(x0)=9πr22L[1+9r24L2+O(rL)4]
To find the orbital period, we consider
(3.1393) L = ( r 0 + ρ ) 2 φ ˙ r 0 2 φ ˙ 1 φ ˙ = r 0 2 L (3.1393) L = r 0 + ρ 2 φ ˙ r 0 2 φ ˙ 1 φ ˙ = r 0 2 L {:(3.1393)L=(r_(0)+rho)^(2)varphi^(˙)≃r_(0)^(2)varphi^(˙)quad=>quad(1)/((varphi^(˙)))=(r_(0)^(2))/(L):}\begin{equation*} L=\left(r_{0}+\rho\right)^{2} \dot{\varphi} \simeq r_{0}^{2} \dot{\varphi} \quad \Rightarrow \quad \frac{1}{\dot{\varphi}}=\frac{r_{0}^{2}}{L} \tag{3.1393} \end{equation*}(3.1393)L=(r0+ρ)2φ˙r02φ˙1φ˙=r02L
assuming that ρ r 0 ρ r 0 rho≪r_(0)\rho \ll r_{0}ρr0. The orbital period is the proper time taken to complete a full turn, i.e.,
(3.1394) T φ = d τ = d φ φ ˙ = 2 π r 0 2 L = 2 π x 0 2 L (3.1394) T φ = d τ = d φ φ ˙ = 2 π r 0 2 L = 2 π x 0 2 L {:(3.1394)T_(varphi)=int d tau=int(d varphi)/((varphi^(˙)))=2pi(r_(0)^(2))/(L)=(2pi)/(x_(0)^(2)L):}\begin{equation*} T_{\varphi}=\int d \tau=\int \frac{d \varphi}{\dot{\varphi}}=2 \pi \frac{r_{0}^{2}}{L}=\frac{2 \pi}{x_{0}^{2} L} \tag{3.1394} \end{equation*}(3.1394)Tφ=dτ=dφφ˙=2πr02L=2πx02L
Inserting the expression for x 0 x 0 x_(0)x_{0}x0, we find that
(3.1395) T φ = 9 π r 2 2 L [ 1 + 3 r 2 2 L 2 + O ( r L ) 4 ] (3.1395) T φ = 9 π r 2 2 L 1 + 3 r 2 2 L 2 + O r L 4 {:(3.1395)T_(varphi)=(9pir_(**)^(2))/(2L)[1+(3r_(**)^(2))/(2L^(2))+O((r_(**))/(L))^(4)]:}\begin{equation*} T_{\varphi}=\frac{9 \pi r_{*}^{2}}{2 L}\left[1+\frac{3 r_{*}^{2}}{2 L^{2}}+\mathcal{O}\left(\frac{r_{*}}{L}\right)^{4}\right] \tag{3.1395} \end{equation*}(3.1395)Tφ=9πr22L[1+3r22L2+O(rL)4]
Thus, finally, the ratio between the periods is
T ρ T φ = 1 + 9 r 2 4 L 2 + O ( r L ) 4 1 + 3 r 2 2 L 2 + O ( r L ) 4 = 1 + ( 9 4 3 2 ) r 2 L 2 + O ( r L ) 4 = 1 + 3 r 2 4 L 2 + O ( r L ) 4 T ρ T φ = 1 + 9 r 2 4 L 2 + O r L 4 1 + 3 r 2 2 L 2 + O r L 4 = 1 + 9 4 3 2 r 2 L 2 + O r L 4 = 1 + 3 r 2 4 L 2 + O r L 4 (T_(rho))/(T_(varphi))=(1+(9r_(**)^(2))/(4L^(2))+O((r_(**))/(L))^(4))/(1+(3r_(**)^(2))/(2L^(2))+O((r_(**))/(L))^(4))=1+((9)/(4)-(3)/(2))(r_(**)^(2))/(L^(2))+O((r_(**))/(L))^(4)=1+(3r_(**)^(2))/(4L^(2))+O((r_(**))/(L))^(4)\frac{T_{\rho}}{T_{\varphi}}=\frac{1+\frac{9 r_{*}^{2}}{4 L^{2}}+\mathcal{O}\left(\frac{r_{*}}{L}\right)^{4}}{1+\frac{3 r_{*}^{2}}{2 L^{2}}+\mathcal{O}\left(\frac{r_{*}}{L}\right)^{4}}=1+\left(\frac{9}{4}-\frac{3}{2}\right) \frac{r_{*}^{2}}{L^{2}}+\mathcal{O}\left(\frac{r_{*}}{L}\right)^{4}=1+\frac{3 r_{*}^{2}}{4 L^{2}}+\mathcal{O}\left(\frac{r_{*}}{L}\right)^{4}TρTφ=1+9r24L2+O(rL)41+3r22L2+O(rL)4=1+(9432)r2L2+O(rL)4=1+3r24L2+O(rL)4.
For r 0 r r 0 r r_(0)≫r_(**)r_{0} \gg r_{*}r0r, we also have
(3.1397) 1 r 0 = ( 1 1 3 r 2 L 2 ) 1 3 r 1 3 r 3 r 2 2 L 2 = r 2 L 2 (3.1397) 1 r 0 = 1 1 3 r 2 L 2 1 3 r 1 3 r 3 r 2 2 L 2 = r 2 L 2 {:(3.1397)(1)/(r_(0))=(1-sqrt(1-(3r_(**)^(2))/(L^(2))))(1)/(3r_(**))≃(1)/(3r_(**))(3r_(**)^(2))/(2L^(2))=(r_(**))/(2L^(2)):}\begin{equation*} \frac{1}{r_{0}}=\left(1-\sqrt{1-\frac{3 r_{*}^{2}}{L^{2}}}\right) \frac{1}{3 r_{*}} \simeq \frac{1}{3 r_{*}} \frac{3 r_{*}^{2}}{2 L^{2}}=\frac{r_{*}}{2 L^{2}} \tag{3.1397} \end{equation*}(3.1397)1r0=(113r2L2)13r13r3r22L2=r2L2
Inserting this into the above leads to
(3.1398) T ρ T φ 1 + 3 r 2 r 0 (3.1398) T ρ T φ 1 + 3 r 2 r 0 {:(3.1398)(T_(rho))/(T_(varphi))≃1+(3r_(**))/(2r_(0)):}\begin{equation*} \frac{T_{\rho}}{T_{\varphi}} \simeq 1+\frac{3 r_{*}}{2 r_{0}} \tag{3.1398} \end{equation*}(3.1398)TρTφ1+3r2r0
2.76
For the past-null geodesics to originate at infinity, it is necessary that there exists a classical turning point. The effective potential for a light signal is given by
(3.1399) V ( r ) = 1 2 L 2 r 2 ( 1 R r ) (3.1399) V ( r ) = 1 2 L 2 r 2 1 R r {:(3.1399)V(r)=(1)/(2)(L^(2))/(r^(2))*(1-(R)/(r)):}\begin{equation*} V(r)=\frac{1}{2} \frac{L^{2}}{r^{2}} \cdot\left(1-\frac{R}{r}\right) \tag{3.1399} \end{equation*}(3.1399)V(r)=12L2r2(1Rr)
For r r r rarr oor \rightarrow \inftyr, we also have V ( r ) 0 V ( r ) 0 V(r)rarr0V(r) \rightarrow 0V(r)0, and consequently, we find that
(3.1400) E = r ˙ 2 2 = 1 2 (3.1400) E = r ˙ 2 2 = 1 2 {:(3.1400)E=(r^(˙)^(2))/(2)=(1)/(2):}\begin{equation*} E=\frac{\dot{r}^{2}}{2}=\frac{1}{2} \tag{3.1400} \end{equation*}(3.1400)E=r˙22=12
when using a parametrization such that r ˙ = 1 r ˙ = 1 r^(˙)=1\dot{r}=1r˙=1 at r r r rarr oor \rightarrow \inftyr. For a classical turning point r r r_(**)r_{*}r to exist, it must hold that the maximum of V ( r ) E V ( r ) E V(r) >= EV(r) \geq EV(r)E for some r r rrr. The extreme points of V V VVV satisfy (with x = 1 / r x = 1 / r x=1//rx=1 / rx=1/r )
(3.1401) V ( r ) = d x d r d d x V ( 1 x ) = d x d r 1 2 L 2 ( 2 3 x R ) x = 0 (3.1401) V ( r ) = d x d r d d x V 1 x = d x d r 1 2 L 2 ( 2 3 x R ) x = 0 {:(3.1401)V^(')(r)=(dx)/(dr)(d)/(dx)V((1)/(x))=(dx)/(dr)(1)/(2)L^(2)(2-3xR)x=0:}\begin{equation*} V^{\prime}(r)=\frac{d x}{d r} \frac{d}{d x} V\left(\frac{1}{x}\right)=\frac{d x}{d r} \frac{1}{2} L^{2}(2-3 x R) x=0 \tag{3.1401} \end{equation*}(3.1401)V(r)=dxdrddxV(1x)=dxdr12L2(23xR)x=0
For r < r < r < oor<\inftyr<, this is solved by r = 3 R 2 r = 3 R 2 r=(3R)/(2)r=\frac{3 R}{2}r=3R2, where the potential obtains a maximum. We find that the condition for a classical turning point to exist is therefore
(3.1402) E = 1 2 V ( 3 R 2 ) = 1 2 4 L 2 9 R 2 ( 1 2 3 ) = 1 2 4 27 L 2 R 2 L 2 27 4 R 2 (3.1402) E = 1 2 V 3 R 2 = 1 2 4 L 2 9 R 2 1 2 3 = 1 2 4 27 L 2 R 2 L 2 27 4 R 2 {:(3.1402)E=(1)/(2) <= V((3R)/(2))=(1)/(2)(4L^(2))/(9R^(2))(1-(2)/(3))=(1)/(2)(4)/(27)(L^(2))/(R^(2))quad=>quadL^(2) <= (27)/(4)R^(2):}\begin{equation*} E=\frac{1}{2} \leq V\left(\frac{3 R}{2}\right)=\frac{1}{2} \frac{4 L^{2}}{9 R^{2}}\left(1-\frac{2}{3}\right)=\frac{1}{2} \frac{4}{27} \frac{L^{2}}{R^{2}} \quad \Rightarrow \quad L^{2} \leq \frac{27}{4} R^{2} \tag{3.1402} \end{equation*}(3.1402)E=12V(3R2)=124L29R2(123)=12427L2R2L2274R2
The smallest impact parameter with a classical turning point is therefore (see Figure 3.13)
(3.1403) b 2 = 27 4 R 2 (3.1403) b 2 = 27 4 R 2 {:(3.1403)b^(2)=(27)/(4)R^(2):}\begin{equation*} b^{2}=\frac{27}{4} R^{2} \tag{3.1403} \end{equation*}(3.1403)b2=274R2
giving an optical size of
(3.1404) 4 π b 2 = 27 π R 2 (3.1404) 4 π b 2 = 27 π R 2 {:(3.1404)4pib^(2)=27 piR^(2):}\begin{equation*} 4 \pi b^{2}=27 \pi R^{2} \tag{3.1404} \end{equation*}(3.1404)4πb2=27πR2
Figure 3.13 Illustration of the optical size of a Schwarzschild black hole ("BH"), where b b bbb is the minimal impact parameter of the optical size 4 π b 2 4 π b 2 4pib^(2)4 \pi b^{2}4πb2.
In other words, note that the optical size of the black hole is significantly larger than 4 π R 2 4 π R 2 4piR^(2)4 \pi R^{2}4πR2.

2.77

Using the given ansatz, we have
(3.1405) d Z 1 = cosh t 2 r f ( r ) d t + 2 r sinh t 2 r f ( r ) d r (3.1406) d Z 2 = sinh t 2 r f ( r ) d t + 2 r cosh t 2 r f ( r ) d r (3.1407) d Z 3 = g ( r ) d r (3.1408) d Z 4 = sin θ cos φ d r + r cos θ cos φ d θ r sin θ sin φ d φ (3.1409) d Z 5 = sin θ sin φ d r + r cos θ sin φ d θ + r sin θ cos φ d φ (3.1410) d Z 6 = cos θ d r r sin θ d θ (3.1405) d Z 1 = cosh t 2 r f ( r ) d t + 2 r sinh t 2 r f ( r ) d r (3.1406) d Z 2 = sinh t 2 r f ( r ) d t + 2 r cosh t 2 r f ( r ) d r (3.1407) d Z 3 = g ( r ) d r (3.1408) d Z 4 = sin θ cos φ d r + r cos θ cos φ d θ r sin θ sin φ d φ (3.1409) d Z 5 = sin θ sin φ d r + r cos θ sin φ d θ + r sin θ cos φ d φ (3.1410) d Z 6 = cos θ d r r sin θ d θ {:[(3.1405)dZ_(1)=cosh((t)/(2r_(**)))f(r)dt+2r_(**)sinh((t)/(2r_(**)))f^(')(r)dr],[(3.1406)dZ_(2)=sinh((t)/(2r_(**)))f(r)dt+2r_(**)cosh((t)/(2r_(**)))f^(')(r)dr],[(3.1407)dZ_(3)=g^(')(r)dr],[(3.1408)dZ_(4)=sin theta cos varphi dr+r cos theta cos varphi d theta-r sin theta sin varphi d varphi],[(3.1409)dZ_(5)=sin theta sin varphi dr+r cos theta sin varphi d theta+r sin theta cos varphi d varphi],[(3.1410)dZ_(6)=cos theta dr-r sin theta d theta]:}\begin{align*} d Z_{1} & =\cosh \frac{t}{2 r_{*}} f(r) d t+2 r_{*} \sinh \frac{t}{2 r_{*}} f^{\prime}(r) d r \tag{3.1405}\\ d Z_{2} & =\sinh \frac{t}{2 r_{*}} f(r) d t+2 r_{*} \cosh \frac{t}{2 r_{*}} f^{\prime}(r) d r \tag{3.1406}\\ d Z_{3} & =g^{\prime}(r) d r \tag{3.1407}\\ d Z_{4} & =\sin \theta \cos \varphi d r+r \cos \theta \cos \varphi d \theta-r \sin \theta \sin \varphi d \varphi \tag{3.1408}\\ d Z_{5} & =\sin \theta \sin \varphi d r+r \cos \theta \sin \varphi d \theta+r \sin \theta \cos \varphi d \varphi \tag{3.1409}\\ d Z_{6} & =\cos \theta d r-r \sin \theta d \theta \tag{3.1410} \end{align*}(3.1405)dZ1=cosht2rf(r)dt+2rsinht2rf(r)dr(3.1406)dZ2=sinht2rf(r)dt+2rcosht2rf(r)dr(3.1407)dZ3=g(r)dr(3.1408)dZ4=sinθcosφdr+rcosθcosφdθrsinθsinφdφ(3.1409)dZ5=sinθsinφdr+rcosθsinφdθ+rsinθcosφdφ(3.1410)dZ6=cosθdrrsinθdθ
Therefore, we find that
(3.1411) d Z 1 2 d Z 2 2 d Z 3 2 = f ( r ) 2 d t 2 [ ( 2 r ) 2 f ( r ) 2 + g ( r ) 2 ] d r 2 (3.1412) d Z 4 2 d Z 5 2 d Z 6 2 = d r 2 r 2 [ d θ 2 + sin ( θ ) 2 d φ 2 ] (3.1411) d Z 1 2 d Z 2 2 d Z 3 2 = f ( r ) 2 d t 2 2 r 2 f ( r ) 2 + g ( r ) 2 d r 2 (3.1412) d Z 4 2 d Z 5 2 d Z 6 2 = d r 2 r 2 d θ 2 + sin ( θ ) 2 d φ 2 {:[(3.1411)dZ_(1)^(2)-dZ_(2)^(2)-dZ_(3)^(2)=f(r)^(2)dt^(2)-[(2r_(**))^(2)f^(')(r)^(2)+g^(')(r)^(2)]dr^(2)],[(3.1412)-dZ_(4)^(2)-dZ_(5)^(2)-dZ_(6)^(2)=-dr^(2)-r^(2)[dtheta^(2)+sin(theta)^(2)dvarphi^(2)]]:}\begin{align*} d Z_{1}^{2}-d Z_{2}^{2}-d Z_{3}^{2} & =f(r)^{2} d t^{2}-\left[\left(2 r_{*}\right)^{2} f^{\prime}(r)^{2}+g^{\prime}(r)^{2}\right] d r^{2} \tag{3.1411}\\ -d Z_{4}^{2}-d Z_{5}^{2}-d Z_{6}^{2} & =-d r^{2}-r^{2}\left[d \theta^{2}+\sin (\theta)^{2} d \varphi^{2}\right] \tag{3.1412} \end{align*}(3.1411)dZ12dZ22dZ32=f(r)2dt2[(2r)2f(r)2+g(r)2]dr2(3.1412)dZ42dZ52dZ62=dr2r2[dθ2+sin(θ)2dφ2]
and thus, the result claimed is equivalent to the following conditions
(3.1413) f ( r ) 2 = 1 r r , ( 2 r ) 2 f ( r ) 2 g ( r ) 2 1 = ( 1 r r ) 1 (3.1413) f ( r ) 2 = 1 r r , 2 r 2 f ( r ) 2 g ( r ) 2 1 = 1 r r 1 {:(3.1413)f(r)^(2)=1-(r_(**))/(r)","quad-(2r_(**))^(2)f^(')(r)^(2)-g^(')(r)^(2)-1=-(1-(r_(**))/(r))^(-1):}\begin{equation*} f(r)^{2}=1-\frac{r_{*}}{r}, \quad-\left(2 r_{*}\right)^{2} f^{\prime}(r)^{2}-g^{\prime}(r)^{2}-1=-\left(1-\frac{r_{*}}{r}\right)^{-1} \tag{3.1413} \end{equation*}(3.1413)f(r)2=1rr,(2r)2f(r)2g(r)21=(1rr)1
After some computations, we obtain
(3.1414) f ( r ) = 1 r r , g ( r ) = r 2 r + r r 2 + r 3 r 3 (3.1414) f ( r ) = 1 r r , g ( r ) = r 2 r + r r 2 + r 3 r 3 {:(3.1414)f(r)=sqrt(1-(r_(**))/(r))","quadg^(')(r)=(r^(2)r_(**)+rr_(**)^(2)+r_(**)^(3))/(r^(3)):}\begin{equation*} f(r)=\sqrt{1-\frac{r_{*}}{r}}, \quad g^{\prime}(r)=\frac{r^{2} r_{*}+r r_{*}^{2}+r_{*}^{3}}{r^{3}} \tag{3.1414} \end{equation*}(3.1414)f(r)=1rr,g(r)=r2r+rr2+r3r3
i.e.,
(3.1415) g ( r ) = ± r 0 r r 2 r + r r 2 + r 3 r 3 d r (3.1415) g ( r ) = ± r 0 r r 2 r + r r 2 + r 3 r 3 d r {:(3.1415)g(r)=+-int_(r_(0))^(r)(r^(2)r_(**)+rr_(**)^(2)+r_(**)^(3))/(r^(3))dr:}\begin{equation*} g(r)= \pm \int_{r_{0}}^{r} \frac{r^{2} r_{*}+r r_{*}^{2}+r_{*}^{3}}{r^{3}} d r \tag{3.1415} \end{equation*}(3.1415)g(r)=±r0rr2r+rr2+r3r3dr
with some arbitrary r 0 > 0 r 0 > 0 r_(0) > 0r_{0}>0r0>0. This proves the claimed result.

2.78

In spherical coordinates, the Minkowski metric is given by
(3.1416) d s 2 = ( d x 0 ) 2 d r 2 r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) (3.1416) d s 2 = d x 0 2 d r 2 r 2 d θ 2 + sin 2 θ d ϕ 2 {:(3.1416)ds^(2)=(dx^(0))^(2)-dr^(2)-r^(2)(dtheta^(2)+sin^(2)theta dphi^(2)):}\begin{equation*} d s^{2}=\left(d x^{0}\right)^{2}-d r^{2}-r^{2}\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right) \tag{3.1416} \end{equation*}(3.1416)ds2=(dx0)2dr2r2(dθ2+sin2θdϕ2)
Using the restriction to the three-dimensional hyperboloid M 3 M 3 M_(3)M_{3}M3, i.e., ( x 0 ) 2 r 2 = x 0 2 r 2 = (x^(0))^(2)-r^(2)=\left(x^{0}\right)^{2}-r^{2}=(x0)2r2= a 2 a 2 -a^(2)-a^{2}a2, we find that
(3.1417) r = ( x 0 ) 2 + a 2 d r = x 0 d x 0 ( x 0 ) 2 + a 2 (3.1417) r = x 0 2 + a 2 d r = x 0 d x 0 x 0 2 + a 2 {:(3.1417)r=sqrt((x^(0))^(2)+a^(2))=>dr=(x^(0)dx^(0))/(sqrt((x^(0))^(2)+a^(2))):}\begin{equation*} r=\sqrt{\left(x^{0}\right)^{2}+a^{2}} \Rightarrow d r=\frac{x^{0} d x^{0}}{\sqrt{\left(x^{0}\right)^{2}+a^{2}}} \tag{3.1417} \end{equation*}(3.1417)r=(x0)2+a2dr=x0dx0(x0)2+a2
Now, inserting the restriction to M 3 M 3 M_(3)M_{3}M3 into the Minkowski metric induces the curved metric on M 3 M 3 M_(3)M_{3}M3, i.e.,
(3.1418) d s 2 = a 2 ( x 0 ) 2 + a 2 ( d x 0 ) 2 [ ( x 0 ) 2 + a 2 ] ( d θ 2 + sin 2 θ d ϕ 2 ) (3.1418) d s 2 = a 2 x 0 2 + a 2 d x 0 2 x 0 2 + a 2 d θ 2 + sin 2 θ d ϕ 2 {:(3.1418)ds^(2)=(a^(2))/((x^(0))^(2)+a^(2))(dx^(0))^(2)-[(x^(0))^(2)+a^(2)](dtheta^(2)+sin^(2)theta dphi^(2)):}\begin{equation*} d s^{2}=\frac{a^{2}}{\left(x^{0}\right)^{2}+a^{2}}\left(d x^{0}\right)^{2}-\left[\left(x^{0}\right)^{2}+a^{2}\right]\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right) \tag{3.1418} \end{equation*}(3.1418)ds2=a2(x0)2+a2(dx0)2[(x0)2+a2](dθ2+sin2θdϕ2)
In the case of lightlike geodesics with d ϕ = 0 d ϕ = 0 d phi=0d \phi=0dϕ=0, since ϕ ϕ phi\phiϕ is assumed to be constant, we have
(3.1419) 0 = d s 2 = a 2 ( x 0 ) 2 + a 2 ( d x 0 ) 2 [ ( x 0 ) 2 + a 2 ] d θ 2 (3.1419) 0 = d s 2 = a 2 x 0 2 + a 2 d x 0 2 x 0 2 + a 2 d θ 2 {:(3.1419)0=ds^(2)=(a^(2))/((x^(0))^(2)+a^(2))(dx^(0))^(2)-[(x^(0))^(2)+a^(2)]dtheta^(2):}\begin{equation*} 0=d s^{2}=\frac{a^{2}}{\left(x^{0}\right)^{2}+a^{2}}\left(d x^{0}\right)^{2}-\left[\left(x^{0}\right)^{2}+a^{2}\right] d \theta^{2} \tag{3.1419} \end{equation*}(3.1419)0=ds2=a2(x0)2+a2(dx0)2[(x0)2+a2]dθ2
Now, we can define the Lagrangian as
(3.1420) L = a 2 ( x 0 ) 2 + a 2 ( x ˙ 0 ) 2 [ ( x 0 ) 2 + a 2 ] θ ˙ 2 (3.1420) L = a 2 x 0 2 + a 2 x ˙ 0 2 x 0 2 + a 2 θ ˙ 2 {:(3.1420)L=(a^(2))/((x^(0))^(2)+a^(2))(x^(˙)^(0))^(2)-[(x^(0))^(2)+a^(2)]theta^(˙)^(2):}\begin{equation*} \mathcal{L}=\frac{a^{2}}{\left(x^{0}\right)^{2}+a^{2}}\left(\dot{x}^{0}\right)^{2}-\left[\left(x^{0}\right)^{2}+a^{2}\right] \dot{\theta}^{2} \tag{3.1420} \end{equation*}(3.1420)L=a2(x0)2+a2(x˙0)2[(x0)2+a2]θ˙2
Using the Euler-Lagrange equation for the θ θ theta\thetaθ-coordinate, we obtain
(3.1421) d d s L θ ˙ L θ = 2 d d s { [ ( x 0 ) 2 + a 2 ] θ ˙ } = 0 (3.1421) d d s L θ ˙ L θ = 2 d d s x 0 2 + a 2 θ ˙ = 0 {:(3.1421)(d)/(ds)(del L)/(del(theta^(˙)))-(del L)/(del theta)=-2(d)/(ds){[(x^(0))^(2)+a^(2)](theta^(˙))}=0:}\begin{equation*} \frac{d}{d s} \frac{\partial L}{\partial \dot{\theta}}-\frac{\partial L}{\partial \theta}=-2 \frac{d}{d s}\left\{\left[\left(x^{0}\right)^{2}+a^{2}\right] \dot{\theta}\right\}=0 \tag{3.1421} \end{equation*}(3.1421)ddsLθ˙Lθ=2dds{[(x0)2+a2]θ˙}=0
which defines a constant of motion that can be immediately integrated to give θ ˙ = A / [ ( x 0 ) 2 + a 2 ] θ ˙ = A / x 0 2 + a 2 theta^(˙)=A//[(x^(0))^(2)+a^(2)]\dot{\theta}=A /\left[\left(x^{0}\right)^{2}+a^{2}\right]θ˙=A/[(x0)2+a2], where A A AAA is some integration constant (in fact, the constant of motion). Furthermore, the condition d s 2 = 0 d s 2 = 0 ds^(2)=0d s^{2}=0ds2=0 implies that
(3.1422) a 2 ( x 0 ) 2 + a 2 ( x ˙ 0 ) 2 [ ( x 0 ) 2 + a 2 ] θ ˙ 2 = 0 (3.1422) a 2 x 0 2 + a 2 x ˙ 0 2 x 0 2 + a 2 θ ˙ 2 = 0 {:(3.1422)(a^(2))/((x^(0))^(2)+a^(2))(x^(˙)^(0))^(2)-[(x^(0))^(2)+a^(2)]theta^(˙)^(2)=0:}\begin{equation*} \frac{a^{2}}{\left(x^{0}\right)^{2}+a^{2}}\left(\dot{x}^{0}\right)^{2}-\left[\left(x^{0}\right)^{2}+a^{2}\right] \dot{\theta}^{2}=0 \tag{3.1422} \end{equation*}(3.1422)a2(x0)2+a2(x˙0)2[(x0)2+a2]θ˙2=0
Hence, combining the two conditions on x ˙ 0 x ˙ 0 x^(˙)^(0)\dot{x}^{0}x˙0 and θ ˙ θ ˙ theta^(˙)\dot{\theta}θ˙ yields
(3.1423) x ˙ 0 = ( x 0 ) 2 + a 2 a θ ˙ = A a x 0 = A a s + α (3.1423) x ˙ 0 = x 0 2 + a 2 a θ ˙ = A a x 0 = A a s + α {:(3.1423)x^(˙)^(0)=((x^(0))^(2)+a^(2))/(a)*theta^(˙)=(A)/(a)quad=>quadx^(0)=(A)/(a)s+alpha:}\begin{equation*} \dot{x}^{0}=\frac{\left(x^{0}\right)^{2}+a^{2}}{a} \cdot \dot{\theta}=\frac{A}{a} \quad \Rightarrow \quad x^{0}=\frac{A}{a} s+\alpha \tag{3.1423} \end{equation*}(3.1423)x˙0=(x0)2+a2aθ˙=Aax0=Aas+α
where α α alpha\alphaα is an integration constant. Thus, inserting the solution for x 0 x 0 x^(0)x^{0}x0 into the Euler-Lagrange equation for the θ θ theta\thetaθ-coordinate, we obtain the lightlike geodesics with d ϕ = 0 d ϕ = 0 d phi=0d \phi=0dϕ=0 as
(3.1424) θ ˙ = A ( x 0 ) 2 + a 2 = A ( A s / a + α ) 2 + a 2 θ = arctan ( A a 2 s + α a ) + β , (3.1424) θ ˙ = A x 0 2 + a 2 = A ( A s / a + α ) 2 + a 2 θ = arctan A a 2 s + α a + β , {:(3.1424)theta^(˙)=(A)/((x^(0))^(2)+a^(2))=(A)/((As//a+alpha)^(2)+a^(2))=>theta=arctan((A)/(a^(2))s+(alpha )/(a))+beta",":}\begin{equation*} \dot{\theta}=\frac{A}{\left(x^{0}\right)^{2}+a^{2}}=\frac{A}{(A s / a+\alpha)^{2}+a^{2}} \Rightarrow \theta=\arctan \left(\frac{A}{a^{2}} s+\frac{\alpha}{a}\right)+\beta, \tag{3.1424} \end{equation*}(3.1424)θ˙=A(x0)2+a2=A(As/a+α)2+a2θ=arctan(Aa2s+αa)+β,
where β β beta\betaβ is an integration constant.

2.79

Introduce the Lagrangian
L = g μ v x ˙ μ x ˙ v = ( x ˙ 0 ) 2 r ˙ 2 r 2 ϕ ˙ 2 = { ( x 0 ) 2 ( x 1 ) 2 ( x 2 ) 2 = 1 } (3.1425) = r 2 r 2 1 r ˙ 2 r ˙ 2 r 2 ϕ ˙ 2 = 1 r 2 1 r ˙ 2 r 2 ϕ ˙ 2 L = g μ v x ˙ μ x ˙ v = x ˙ 0 2 r ˙ 2 r 2 ϕ ˙ 2 = x 0 2 x 1 2 x 2 2 = 1 (3.1425) = r 2 r 2 1 r ˙ 2 r ˙ 2 r 2 ϕ ˙ 2 = 1 r 2 1 r ˙ 2 r 2 ϕ ˙ 2 {:[L=g_(mu v)x^(˙)^(mu)x^(˙)^(v)=(x^(˙)^(0))^(2)-r^(˙)^(2)-r^(2)phi^(˙)^(2)={(x^(0))^(2)-(x^(1))^(2)-(x^(2))^(2)=-1}],[(3.1425)=(r^(2))/(r^(2)-1)r^(˙)^(2)-r^(˙)^(2)-r^(2)phi^(˙)^(2)=(1)/(r^(2)-1)*r^(˙)^(2)-r^(2)phi^(˙)^(2)]:}\begin{align*} \mathscr{L} & =g_{\mu v} \dot{x}^{\mu} \dot{x}^{v}=\left(\dot{x}^{0}\right)^{2}-\dot{r}^{2}-r^{2} \dot{\phi}^{2}=\left\{\left(x^{0}\right)^{2}-\left(x^{1}\right)^{2}-\left(x^{2}\right)^{2}=-1\right\} \\ & =\frac{r^{2}}{r^{2}-1} \dot{r}^{2}-\dot{r}^{2}-r^{2} \dot{\phi}^{2}=\frac{1}{r^{2}-1} \cdot \dot{r}^{2}-r^{2} \dot{\phi}^{2} \tag{3.1425} \end{align*}L=gμvx˙μx˙v=(x˙0)2r˙2r2ϕ˙2={(x0)2(x1)2(x2)2=1}(3.1425)=r2r21r˙2r˙2r2ϕ˙2=1r21r˙2r2ϕ˙2
where it holds that x 0 = r 2 1 x 0 = r 2 1 x^(0)=sqrt(r^(2)-1)x^{0}=\sqrt{r^{2}-1}x0=r21. The Euler-Lagrange variational equations, i.e., d d s ( L x ˙ μ ) L x μ = 0 d d s L x ˙ μ L x μ = 0 (d)/(ds)((delL)/(delx^(˙)^(mu)))-(delL)/(delx^(mu))=0\frac{d}{d s}\left(\frac{\partial \mathscr{L}}{\partial \dot{x}^{\mu}}\right)-\frac{\partial \mathscr{L}}{\partial x^{\mu}}=0dds(Lx˙μ)Lxμ=0, then become
(3.1426) d d s ( r ˙ r 2 1 ) + r ϕ ˙ 2 + r r ˙ 2 ( r 2 1 ) 2 = 0 for r (3.1427) d d s ( r 2 ϕ ˙ ) = 0 for ϕ (3.1426) d d s r ˙ r 2 1 + r ϕ ˙ 2 + r r ˙ 2 r 2 1 2 = 0  for  r (3.1427) d d s r 2 ϕ ˙ = 0  for  ϕ {:[(3.1426)(d)/(ds)(((r^(˙)))/(r^(2)-1))+rphi^(˙)^(2)+(rr^(˙)^(2))/((r^(2)-1)^(2))=0" for "r],[(3.1427)(d)/(ds)(r^(2)(phi^(˙)))=0" for "phi]:}\begin{align*} \frac{d}{d s}\left(\frac{\dot{r}}{r^{2}-1}\right)+r \dot{\phi}^{2}+\frac{r \dot{r}^{2}}{\left(r^{2}-1\right)^{2}} & =0 \text { for } r \tag{3.1426}\\ \frac{d}{d s}\left(r^{2} \dot{\phi}\right) & =0 \text { for } \phi \tag{3.1427} \end{align*}(3.1426)dds(r˙r21)+rϕ˙2+rr˙2(r21)2=0 for r(3.1427)dds(r2ϕ˙)=0 for ϕ
Furthermore, along a lightlike curve L = 0 L = 0 L=0\mathscr{L}=0L=0, and thus, we have
(3.1428) 1 r 2 1 r ˙ 2 r 2 ϕ ˙ 2 = 0 (3.1428) 1 r 2 1 r ˙ 2 r 2 ϕ ˙ 2 = 0 {:(3.1428)(1)/(r^(2)-1)r^(˙)^(2)-r^(2)phi^(˙)^(2)=0:}\begin{equation*} \frac{1}{r^{2}-1} \dot{r}^{2}-r^{2} \dot{\phi}^{2}=0 \tag{3.1428} \end{equation*}(3.1428)1r21r˙2r2ϕ˙2=0
From the Euler-Lagrange variational equation for ϕ ϕ phi\phiϕ, we find that r 2 ϕ ˙ = A = r 2 ϕ ˙ = A = r^(2)phi^(˙)=A=r^{2} \dot{\phi}=A=r2ϕ˙=A= const. Inserting this into the lightlike condition, we obtain
(3.1429) 1 r 2 1 r ˙ 2 A 2 r 2 = 0 d r d s = r ˙ = A r r 2 1 d d s r 2 1 = A (3.1429) 1 r 2 1 r ˙ 2 A 2 r 2 = 0 d r d s = r ˙ = A r r 2 1 d d s r 2 1 = A {:(3.1429)(1)/(r^(2)-1)r^(˙)^(2)-(A^(2))/(r^(2))=0=>(dr)/(ds)=r^(˙)=(A)/(r)sqrt(r^(2)-1)quad=>quad(d)/(ds)sqrt(r^(2)-1)=A:}\begin{equation*} \frac{1}{r^{2}-1} \dot{r}^{2}-\frac{A^{2}}{r^{2}}=0 \Rightarrow \frac{d r}{d s}=\dot{r}=\frac{A}{r} \sqrt{r^{2}-1} \quad \Rightarrow \quad \frac{d}{d s} \sqrt{r^{2}-1}=A \tag{3.1429} \end{equation*}(3.1429)1r21r˙2A2r2=0drds=r˙=Arr21ddsr21=A
so that r 2 1 = A s + s 0 r 2 1 = A s + s 0 sqrt(r^(2)-1)=As+s_(0)\sqrt{r^{2}-1}=A s+s_{0}r21=As+s0 or r = 1 + ( A s + s 0 ) 2 r = 1 + A s + s 0 2 r=sqrt(1+(As+s_(0))^(2))r=\sqrt{1+\left(A s+s_{0}\right)^{2}}r=1+(As+s0)2. From r 2 ϕ ˙ = A r 2 ϕ ˙ = A r^(2)phi^(˙)=Ar^{2} \dot{\phi}=Ar2ϕ˙=A, we find that
(3.1430) d ϕ d s = ϕ ˙ = A 1 + ( A s + s 0 ) 2 (3.1430) d ϕ d s = ϕ ˙ = A 1 + A s + s 0 2 {:(3.1430)(d phi)/(ds)=phi^(˙)=(A)/(1+(As+s_(0))^(2)):}\begin{equation*} \frac{d \phi}{d s}=\dot{\phi}=\frac{A}{1+\left(A s+s_{0}\right)^{2}} \tag{3.1430} \end{equation*}(3.1430)dϕds=ϕ˙=A1+(As+s0)2
which implies that
(3.1431) ϕ = ϕ 0 + arctan ( A s + s 0 ) ϕ 0 + arctan s ~ (3.1431) ϕ = ϕ 0 + arctan A s + s 0 ϕ 0 + arctan s ~ {:(3.1431)phi=phi_(0)+arctan(As+s_(0))-=phi_(0)+arctan tilde(s):}\begin{equation*} \phi=\phi_{0}+\arctan \left(A s+s_{0}\right) \equiv \phi_{0}+\arctan \tilde{s} \tag{3.1431} \end{equation*}(3.1431)ϕ=ϕ0+arctan(As+s0)ϕ0+arctans~
Now, x 0 = r 2 1 = A s + s 0 = s ~ = tan ( ϕ ϕ 0 ) x 0 = r 2 1 = A s + s 0 = s ~ = tan ϕ ϕ 0 x^(0)=sqrt(r^(2)-1)=As+s_(0)= tilde(s)=tan(phi-phi_(0))x^{0}=\sqrt{r^{2}-1}=A s+s_{0}=\tilde{s}=\tan \left(\phi-\phi_{0}\right)x0=r21=As+s0=s~=tan(ϕϕ0), i.e., x 0 = tan ( ϕ ϕ 0 ) x 0 = tan ϕ ϕ 0 x^(0)=tan(phi-phi_(0))x^{0}=\tan \left(\phi-\phi_{0}\right)x0=tan(ϕϕ0), and finally, Δ ϕ ϕ ϕ 0 = π 2 Δ ϕ ϕ ϕ 0 = π 2 Delta phi-=phi-phi_(0)=(pi)/(2)\Delta \phi \equiv \phi-\phi_{0}=\frac{\pi}{2}Δϕϕϕ0=π2, which corresponds to Δ x 0 = tan Δ ϕ Δ x 0 = tan Δ ϕ Deltax^(0)=tan Delta phi rarr oo\Delta x^{0}=\tan \Delta \phi \rightarrow \inftyΔx0=tanΔϕ. Thus, it takes an infinite global time difference Δ x 0 Δ x 0 Deltax^(0)\Delta x^{0}Δx0 for a light signal to travel from the point ϕ 0 = 0 ϕ 0 = 0 phi_(0)=0\phi_{0}=0ϕ0=0 to the point ϕ = π 2 ϕ = π 2 phi=(pi)/(2)\phi=\frac{\pi}{2}ϕ=π2. on the surface.
2.80
The Schwarzschild metric (for θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2 ) is given by
(3.1432) d s 2 = α ( d x 0 ) 2 α 1 d r 2 r 2 d ϕ 2 (3.1432) d s 2 = α d x 0 2 α 1 d r 2 r 2 d ϕ 2 {:(3.1432)ds^(2)=alpha(dx^(0))^(2)-alpha^(-1)dr^(2)-r^(2)dphi^(2):}\begin{equation*} d s^{2}=\alpha\left(d x^{0}\right)^{2}-\alpha^{-1} d r^{2}-r^{2} d \phi^{2} \tag{3.1432} \end{equation*}(3.1432)ds2=α(dx0)2α1dr2r2dϕ2
where α α ( r ) = 1 2 G M r α α ( r ) = 1 2 G M r alpha-=alpha(r)=1-(2GM)/(r)\alpha \equiv \alpha(r)=1-\frac{2 G M}{r}αα(r)=12GMr. The geodesic equations are derived by varying L = g μ ν x ˙ μ x ˙ ν L = g μ ν x ˙ μ x ˙ ν L=g_(mu nu)x^(˙)^(mu)x^(˙)^(nu)\mathcal{L}=g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}L=gμνx˙μx˙ν, and therefore, using Euler-Lagrange equations for the x 0 x 0 x^(0)x^{0}x0 and ϕ ϕ phi\phiϕ coordinates, we obtain
(3.1433) x ¨ 0 + 1 α d α d r x ˙ 0 r ˙ = 0 (3.1434) ϕ ¨ + 2 r r ˙ ϕ ˙ = 0 (3.1433) x ¨ 0 + 1 α d α d r x ˙ 0 r ˙ = 0 (3.1434) ϕ ¨ + 2 r r ˙ ϕ ˙ = 0 {:[(3.1433)x^(¨)^(0)+(1)/(alpha)(d alpha)/(dr)x^(˙)^(0)r^(˙)=0],[(3.1434)phi^(¨)+(2)/(r)*r^(˙)phi^(˙)=0]:}\begin{align*} \ddot{x}^{0}+\frac{1}{\alpha} \frac{d \alpha}{d r} \dot{x}^{0} \dot{r} & =0 \tag{3.1433}\\ \ddot{\phi}+\frac{2}{r} \cdot \dot{r} \dot{\phi} & =0 \tag{3.1434} \end{align*}(3.1433)x¨0+1αdαdrx˙0r˙=0(3.1434)ϕ¨+2rr˙ϕ˙=0
which are the geodesic equations x ¨ λ + Γ μ ν λ x ˙ μ x ˙ ν = 0 x ¨ λ + Γ μ ν λ x ˙ μ x ˙ ν = 0 x^(¨)^(lambda)+Gamma_(mu nu)^(lambda)x^(˙)^(mu)x^(˙)^(nu)=0\ddot{x}^{\lambda}+\Gamma_{\mu \nu}^{\lambda} \dot{x}^{\mu} \dot{x}^{\nu}=0x¨λ+Γμνλx˙μx˙ν=0 (see also Problem 2.70). In this case, we actually have the simpler geodesic equations, namely
(3.1435) α x ˙ 0 = E , (3.1436) r 2 ϕ ˙ = h , (3.1435) α x ˙ 0 = E , (3.1436) r 2 ϕ ˙ = h , {:[(3.1435)alphax^(˙)^(0)=E","],[(3.1436)r^(2)phi^(˙)=h","]:}\begin{align*} \alpha \dot{x}^{0} & =E, \tag{3.1435}\\ r^{2} \dot{\phi} & =h, \tag{3.1436} \end{align*}(3.1435)αx˙0=E,(3.1436)r2ϕ˙=h,
where E E EEE and h h hhh are constants. For a lightlike geodesic ( L = 0 ) ( L = 0 ) (L=0)(\mathcal{L}=0)(L=0) in the plane θ = π 2 θ = π 2 theta=(pi)/(2)\theta=\frac{\pi}{2}θ=π2, we have
(3.1437) 0 = g 00 ( x ˙ 0 ) 2 + g r r r ˙ 2 + g ϕ ϕ ϕ ˙ 2 (3.1437) 0 = g 00 x ˙ 0 2 + g r r r ˙ 2 + g ϕ ϕ ϕ ˙ 2 {:(3.1437)0=g_(00)(x^(˙)^(0))^(2)+g_(rr)r^(˙)^(2)+g_(phi phi)phi^(˙)^(2):}\begin{equation*} 0=g_{00}\left(\dot{x}^{0}\right)^{2}+g_{r r} \dot{r}^{2}+g_{\phi \phi} \dot{\phi}^{2} \tag{3.1437} \end{equation*}(3.1437)0=g00(x˙0)2+grrr˙2+gϕϕϕ˙2
where g 00 = α , g r r = 1 α g 00 = α , g r r = 1 α g_(00)=alpha,g_(rr)=-(1)/(alpha)g_{00}=\alpha, g_{r r}=-\frac{1}{\alpha}g00=α,grr=1α, and g ϕ ϕ = r 2 g ϕ ϕ = r 2 g_(phi phi)=-r^(2)g_{\phi \phi}=-r^{2}gϕϕ=r2. Now, the geodesic equation for x 0 x 0 x^(0)x^{0}x0 implies that x ˙ 0 = E α x ˙ 0 = E α x^(˙)^(0)=(E)/( alpha)\dot{x}^{0}=\frac{E}{\alpha}x˙0=Eα, where E E EEE is a constant of motion, whereas the geodesic equation for ϕ ϕ phi\phiϕ implies that ϕ ˙ = h r 2 ϕ ˙ = h r 2 phi^(˙)=(h)/(r^(2))\dot{\phi}=\frac{h}{r^{2}}ϕ˙=hr2, where h h hhh is another constant of motion. Inserting the two constants of motion into the condition for a lightlike geodesic yields
(3.1438) 0 = g 00 E 2 α 2 + g r r r ˙ 2 + g ϕ ϕ h 2 r 4 (3.1438) 0 = g 00 E 2 α 2 + g r r r ˙ 2 + g ϕ ϕ h 2 r 4 {:(3.1438)0=g_(00)(E^(2))/(alpha^(2))+g_(rr)r^(˙)^(2)+g_(phi phi)(h^(2))/(r^(4)):}\begin{equation*} 0=g_{00} \frac{E^{2}}{\alpha^{2}}+g_{r r} \dot{r}^{2}+g_{\phi \phi} \frac{h^{2}}{r^{4}} \tag{3.1438} \end{equation*}(3.1438)0=g00E2α2+grrr˙2+gϕϕh2r4
Solving the above equation for r ˙ r ˙ r^(˙)\dot{r}r˙ gives
(3.1439) r ˙ = g 00 g r r E 2 α 2 g ϕ ϕ g r r h 2 r 4 = E 2 α h 2 r 2 (3.1439) r ˙ = g 00 g r r E 2 α 2 g ϕ ϕ g r r h 2 r 4 = E 2 α h 2 r 2 {:(3.1439)r^(˙)=sqrt(-(g_(00))/(g_(rr))(E^(2))/(alpha^(2))-(g_(phi phi))/(g_(rr))(h^(2))/(r^(4)))=sqrt(E^(2)-alpha(h^(2))/(r^(2))):}\begin{equation*} \dot{r}=\sqrt{-\frac{g_{00}}{g_{r r}} \frac{E^{2}}{\alpha^{2}}-\frac{g_{\phi \phi}}{g_{r r}} \frac{h^{2}}{r^{4}}}=\sqrt{E^{2}-\alpha \frac{h^{2}}{r^{2}}} \tag{3.1439} \end{equation*}(3.1439)r˙=g00grrE2α2gϕϕgrrh2r4=E2αh2r2
Thus, using the fact that r ˙ = d r d s = f ( r ) r ˙ = d r d s = f ( r ) r^(˙)=(dr)/(ds)=f(r)\dot{r}=\frac{d r}{d s}=f(r)r˙=drds=f(r) and α = 1 2 G M r α = 1 2 G M r alpha=1-(2GM)/(r)\alpha=1-\frac{2 G M}{r}α=12GMr, we obtain the answer
(3.1440) d r d s = f ( r ) = E 2 ( 1 2 G M r ) h 2 r 2 (3.1440) d r d s = f ( r ) = E 2 1 2 G M r h 2 r 2 {:(3.1440)(dr)/(ds)=f(r)=sqrt(E^(2)-(1-(2GM)/(r))(h^(2))/(r^(2))):}\begin{equation*} \frac{d r}{d s}=f(r)=\sqrt{E^{2}-\left(1-\frac{2 G M}{r}\right) \frac{h^{2}}{r^{2}}} \tag{3.1440} \end{equation*}(3.1440)drds=f(r)=E2(12GMr)h2r2
where E E EEE and h h hhh are constants of motion, which is the sought-after differential equation for r ( s ) r ( s ) r(s)r(s)r(s) when restricted to the plane θ = π 2 θ = π 2 theta=(pi)/(2)\theta=\frac{\pi}{2}θ=π2.

2.81

Consider the metric
(3.1441) d s 2 = g μ ν d x μ d x ν = c 2 d t 2 S ( t ) 2 ( d x 2 + d y 2 + d z 2 ) (3.1441) d s 2 = g μ ν d x μ d x ν = c 2 d t 2 S ( t ) 2 d x 2 + d y 2 + d z 2 {:(3.1441)ds^(2)=g_(mu nu)dx^(mu)dx^(nu)=c^(2)dt^(2)-S(t)^(2)(dx^(2)+dy^(2)+dz^(2)):}\begin{equation*} d s^{2}=g_{\mu \nu} d x^{\mu} d x^{\nu}=c^{2} d t^{2}-S(t)^{2}\left(d x^{2}+d y^{2}+d z^{2}\right) \tag{3.1441} \end{equation*}(3.1441)ds2=gμνdxμdxν=c2dt2S(t)2(dx2+dy2+dz2)
where S ( t ) S ( t ) S(t)S(t)S(t) is an increasing function of time t t ttt with S ( 0 ) = 0 S ( 0 ) = 0 S(0)=0S(0)=0S(0)=0. To find the geodesic equations of motion, we vary the action
(3.1442) S = L d s (3.1442) S = L d s {:(3.1442)S=intLds:}\begin{equation*} S=\int \mathcal{L} d s \tag{3.1442} \end{equation*}(3.1442)S=Lds
where the Lagrangian is given by
(3.1443) L = g μ ν x ˙ μ x ˙ ν = c 2 t ˙ 2 S ( t ) 2 ( x ˙ 2 + y ˙ 2 + z ˙ 2 ) (3.1443) L = g μ ν x ˙ μ x ˙ ν = c 2 t ˙ 2 S ( t ) 2 x ˙ 2 + y ˙ 2 + z ˙ 2 {:(3.1443)L=g_(mu nu)x^(˙)^(mu)x^(˙)^(nu)=c^(2)t^(˙)^(2)-S(t)^(2)(x^(˙)^(2)+y^(˙)^(2)+z^(˙)^(2)):}\begin{equation*} \mathcal{L}=g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=c^{2} \dot{t}^{2}-S(t)^{2}\left(\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2}\right) \tag{3.1443} \end{equation*}(3.1443)L=gμνx˙μx˙ν=c2t˙2S(t)2(x˙2+y˙2+z˙2)
The Euler-Lagrange equations are
(3.1444) d d s L x ˙ μ L x μ = 0 (3.1444) d d s L x ˙ μ L x μ = 0 {:(3.1444)(d)/(ds)(delL)/(delx^(˙)^(mu))-(delL)/(delx^(mu))=0:}\begin{equation*} \frac{d}{d s} \frac{\partial \mathcal{L}}{\partial \dot{x}^{\mu}}-\frac{\partial \mathcal{L}}{\partial x^{\mu}}=0 \tag{3.1444} \end{equation*}(3.1444)ddsLx˙μLxμ=0
which give the following geodesic equations
(3.1445) t ¨ + 1 c 2 S ( t ) S ( t ) ( x ˙ 2 + y ˙ 2 + z ˙ 2 ) = 0 , (3.1446) x ¨ + 2 S ( t ) 1 S ( t ) t ˙ x ˙ = 0 , (3.1447) y ¨ + 2 S ( t ) 1 S ( t ) t ˙ y ˙ = 0 (3.1448) z ¨ + 2 S ( t ) 1 S ( t ) t ˙ z ˙ = 0 (3.1445) t ¨ + 1 c 2 S ( t ) S ( t ) x ˙ 2 + y ˙ 2 + z ˙ 2 = 0 , (3.1446) x ¨ + 2 S ( t ) 1 S ( t ) t ˙ x ˙ = 0 , (3.1447) y ¨ + 2 S ( t ) 1 S ( t ) t ˙ y ˙ = 0 (3.1448) z ¨ + 2 S ( t ) 1 S ( t ) t ˙ z ˙ = 0 {:[(3.1445)t^(¨)+(1)/(c^(2))S(t)S^(')(t)(x^(˙)^(2)+y^(˙)^(2)+z^(˙)^(2))=0","],[(3.1446)x^(¨)+2S(t)^(-1)S^(')(t)t^(˙)x^(˙)=0","],[(3.1447)y^(¨)+2S(t)^(-1)S^(')(t)t^(˙)y^(˙)=0],[(3.1448)z^(¨)+2S(t)^(-1)S^(')(t)t^(˙)z^(˙)=0]:}\begin{align*} \ddot{t}+\frac{1}{c^{2}} S(t) S^{\prime}(t)\left(\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2}\right) & =0, \tag{3.1445}\\ \ddot{x}+2 S(t)^{-1} S^{\prime}(t) \dot{t} \dot{x} & =0, \tag{3.1446}\\ \ddot{y}+2 S(t)^{-1} S^{\prime}(t) \dot{t} \dot{y} & =0 \tag{3.1447}\\ \ddot{z}+2 S(t)^{-1} S^{\prime}(t) \dot{t} \dot{z} & =0 \tag{3.1448} \end{align*}(3.1445)t¨+1c2S(t)S(t)(x˙2+y˙2+z˙2)=0,(3.1446)x¨+2S(t)1S(t)t˙x˙=0,(3.1447)y¨+2S(t)1S(t)t˙y˙=0(3.1448)z¨+2S(t)1S(t)t˙z˙=0
Now, for lightlike geodesics, we have d s = 0 d s = 0 ds=0d s=0ds=0. Note that we may always rotate a coordinate system such that the motion is directed along one spatial coordinate only (say x x xxx ) and the motion is taking place in the positive direction. Then, inserting d s = 0 d s = 0 ds=0d s=0ds=0 and the given condition S ( t ) = t / t 0 S ( t ) = t / t 0 S(t)=t//t_(0)S(t)=t / t_{0}S(t)=t/t0 for t 0 > 0 t 0 > 0 t_(0) > 0t_{0}>0t0>0 into the metric, we find that
(3.1449) 0 = c 2 d t 2 ( t t 0 ) 2 d x 2 c d t = + t t 0 d x d x = c t 0 t d t (3.1449) 0 = c 2 d t 2 t t 0 2 d x 2 c d t = + t t 0 d x d x = c t 0 t d t {:(3.1449)0=c^(2)dt^(2)-((t)/(t_(0)))^(2)dx^(2)=>cdt=+(t)/(t_(0))dx quad=>quad dx=(ct_(0))/(t)dt:}\begin{equation*} 0=c^{2} d t^{2}-\left(\frac{t}{t_{0}}\right)^{2} d x^{2} \Rightarrow c d t=+\frac{t}{t_{0}} d x \quad \Rightarrow \quad d x=\frac{c t_{0}}{t} d t \tag{3.1449} \end{equation*}(3.1449)0=c2dt2(tt0)2dx2cdt=+tt0dxdx=ct0tdt
Integrating this equation yields
(3.1450) x = c t 0 ln ( t / τ ) (3.1450) x = c t 0 ln ( t / τ ) {:(3.1450)x=ct_(0)ln(t//tau):}\begin{equation*} x=c t_{0} \ln (t / \tau) \tag{3.1450} \end{equation*}(3.1450)x=ct0ln(t/τ)
where τ τ tau\tauτ is some constant that can be determined from the initial conditions. Thus, the lightlike geodesics are given by
(3.1451) x = c t 0 ln ( t / τ ) e r + k (3.1451) x = c t 0 ln ( t / τ ) e r + k {:(3.1451)x=ct_(0)ln(t//tau)e_(r)+k:}\begin{equation*} \boldsymbol{x}=c t_{0} \ln (t / \tau) \boldsymbol{e}_{r}+\boldsymbol{k} \tag{3.1451} \end{equation*}(3.1451)x=ct0ln(t/τ)er+k
where e r e r e_(r)e_{r}er is the initial direction of motion and k k kkk is a constant.
Finally, for the event (or spacetime point) p = ( c t 0 , c t 0 , 0 , 0 ) p = c t 0 , c t 0 , 0 , 0 p=(ct_(0),ct_(0),0,0)p=\left(c t_{0}, c t_{0}, 0,0\right)p=(ct0,ct0,0,0), the points on the future light cone are given by
(3.1452) x = c t 0 ln ( t / t 0 ) e r + k (3.1452) x = c t 0 ln t / t 0 e r + k {:(3.1452)x=ct_(0)ln(t//t_(0))e_(r)+k:}\begin{equation*} \boldsymbol{x}=c t_{0} \ln \left(t / t_{0}\right) \boldsymbol{e}_{r}+\boldsymbol{k} \tag{3.1452} \end{equation*}(3.1452)x=ct0ln(t/t0)er+k
where k = ( c t 0 , 0 , 0 ) k = c t 0 , 0 , 0 k=(ct_(0),0,0)k=\left(c t_{0}, 0,0\right)k=(ct0,0,0). Therefore, the set of events J + ( p ) J + ( p ) J^(+)(p)J^{+}(p)J+(p) that are causally connected to p p ppp are the points inside this light cone, i.e.,
(3.1453) J + ( p ) = { ( c t , x , y , z ) R 4 : ( x c t 0 ) 2 + y 2 + z 2 [ c t 0 ln ( t / t 0 ) ] 2 } (3.1453) J + ( p ) = ( c t , x , y , z ) R 4 : x c t 0 2 + y 2 + z 2 c t 0 ln t / t 0 2 {:(3.1453)J^(+)(p)={(ct,x,y,z)inR^(4):(x-ct_(0))^(2)+y^(2)+z^(2) <= [ct_(0)ln(t//t_(0))]^(2)}:}\begin{equation*} J^{+}(p)=\left\{(c t, x, y, z) \in \mathbb{R}^{4}:\left(x-c t_{0}\right)^{2}+y^{2}+z^{2} \leq\left[c t_{0} \ln \left(t / t_{0}\right)\right]^{2}\right\} \tag{3.1453} \end{equation*}(3.1453)J+(p)={(ct,x,y,z)R4:(xct0)2+y2+z2[ct0ln(t/t0)]2}
2.82
From the setup of the problem (see Figure 3.14), we have (at the equator θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2 ):
(3.1454) θ ˙ = 1 , φ ˙ = 0 , for the geodesics, (3.1455) X θ = 0 , X φ = δ , for the separation. (3.1454) θ ˙ = 1 , φ ˙ = 0 ,  for the geodesics,  (3.1455) X θ = 0 , X φ = δ ,  for the separation.  {:[(3.1454)theta^(˙)=1","varphi^(˙)=0","" for the geodesics, "],[(3.1455)X^(theta)=0","X^(varphi)=delta","" for the separation. "]:}\begin{align*} \dot{\theta}=1, & \dot{\varphi}=0, & & \text { for the geodesics, } \tag{3.1454}\\ X^{\theta}=0, & X^{\varphi}=\delta, & & \text { for the separation. } \tag{3.1455} \end{align*}(3.1454)θ˙=1,φ˙=0, for the geodesics, (3.1455)Xθ=0,Xφ=δ, for the separation. 
It follows that
(3.1456) A θ = R θ θ φ θ δ , A φ = R θ θ φ φ δ (3.1456) A θ = R θ θ φ θ δ , A φ = R θ θ φ φ δ {:(3.1456)A^(theta)=R_(theta theta varphi)^(theta)delta","quadA^(varphi)=R_(theta theta varphi)^(varphi)delta:}\begin{equation*} A^{\theta}=R_{\theta \theta \varphi}^{\theta} \delta, \quad A^{\varphi}=R_{\theta \theta \varphi}^{\varphi} \delta \tag{3.1456} \end{equation*}(3.1456)Aθ=Rθθφθδ,Aφ=Rθθφφδ
Figure 3.14 Illustration of the setup of the problem. On the unit sphere S 2 S 2 S^(2)\mathbb{S}^{2}S2, two geodesics are separated by a small distance δ δ delta\deltaδ and both are orthogonal to the equator θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2.
Noting that R θ θ θ φ = g θ θ R θ θ θ φ = 0 R θ θ θ φ = g θ θ R θ θ θ φ = 0 R^(theta)_(theta theta varphi)=g^(theta theta)R_(theta theta theta varphi)=0R^{\theta}{ }_{\theta \theta \varphi}=g^{\theta \theta} R_{\theta \theta \theta \varphi}=0Rθθθφ=gθθRθθθφ=0, it thus follows that A θ = 0 A θ = 0 A^(theta)=0A^{\theta}=0Aθ=0. For R θ θ φ φ R θ θ φ φ R_(theta theta varphi)^(varphi)R_{\theta \theta \varphi}^{\varphi}Rθθφφ, we find that
R θ θ φ a a = θ φ θ φ θ θ = θ ( Γ φ θ b b ) φ ( Γ θ θ b b ) = θ ( Γ φ θ b b ) 0 = θ ( cos θ sin θ φ ) = ( sin θ sin θ cos 2 θ sin 2 θ ) φ + cot θ Γ θ φ b b (3.1457) = ( 1 cot 2 θ + cot 2 θ ) φ = φ R θ θ φ a a = θ φ θ φ θ θ = θ Γ φ θ b b φ Γ θ θ b b = θ Γ φ θ b b 0 = θ cos θ sin θ φ = sin θ sin θ cos 2 θ sin 2 θ φ + cot θ Γ θ φ b b (3.1457) = 1 cot 2 θ + cot 2 θ φ = φ {:[R_(theta theta varphi)^(a)del_(a)=grad_(theta)grad_(varphi)del_(theta)-grad_(varphi)grad_(theta)del_(theta)=grad_(theta)(Gamma_(varphi theta)^(b)del_(b))-grad_(varphi)(Gamma_(theta theta)^(b)del_(b))=grad_(theta)(Gamma_(varphi theta)^(b)del_(b))-0],[=grad_(theta)((cos theta)/(sin theta)del_(varphi))=(-(sin theta)/(sin theta)-(cos^(2)theta)/(sin^(2)theta))del_(varphi)+cot thetaGamma_(theta varphi)^(b)del_(b)],[(3.1457)=(-1-cot^(2)theta+cot^(2)theta)del_(varphi)=-del_(varphi)]:}\begin{align*} R_{\theta \theta \varphi}^{a} \partial_{a} & =\nabla_{\theta} \nabla_{\varphi} \partial_{\theta}-\nabla_{\varphi} \nabla_{\theta} \partial_{\theta}=\nabla_{\theta}\left(\Gamma_{\varphi \theta}^{b} \partial_{b}\right)-\nabla_{\varphi}\left(\Gamma_{\theta \theta}^{b} \partial_{b}\right)=\nabla_{\theta}\left(\Gamma_{\varphi \theta}^{b} \partial_{b}\right)-0 \\ & =\nabla_{\theta}\left(\frac{\cos \theta}{\sin \theta} \partial_{\varphi}\right)=\left(-\frac{\sin \theta}{\sin \theta}-\frac{\cos ^{2} \theta}{\sin ^{2} \theta}\right) \partial_{\varphi}+\cot \theta \Gamma_{\theta \varphi}^{b} \partial_{b} \\ & =\left(-1-\cot ^{2} \theta+\cot ^{2} \theta\right) \partial_{\varphi}=-\partial_{\varphi} \tag{3.1457} \end{align*}Rθθφaa=θφθφθθ=θ(Γφθbb)φ(Γθθbb)=θ(Γφθbb)0=θ(cosθsinθφ)=(sinθsinθcos2θsin2θ)φ+cotθΓθφbb(3.1457)=(1cot2θ+cot2θ)φ=φ
which implies that R θ θ φ φ = 1 R θ θ φ φ = 1 R_(theta theta varphi)^(varphi)=-1R_{\theta \theta \varphi}^{\varphi}=-1Rθθφφ=1. We therefore conclude that A φ = δ A φ = δ A^(varphi)=-deltaA^{\varphi}=-\deltaAφ=δ.

2.83

a) The trajectories x μ ( τ ) x μ ( τ ) x^(mu)(tau)x^{\mu}(\tau)xμ(τ) of freely falling particles or photons are the geodesics of the given metric and can be obtained from the variational principle
(3.1458) δ 1 2 s ˙ 2 d τ = δ 1 2 [ t ˙ 2 e 2 t / R ( x ˙ 2 + y ˙ 2 + z ˙ 2 ) ] d τ = 0 (3.1458) δ 1 2 s ˙ 2 d τ = δ 1 2 t ˙ 2 e 2 t / R x ˙ 2 + y ˙ 2 + z ˙ 2 d τ = 0 {:(3.1458)delta int(1)/(2)s^(˙)^(2)d tau=delta int(1)/(2)[t^(˙)^(2)-e^(2t//R)(x^(˙)^(2)+y^(˙)^(2)+z^(˙)^(2))]d tau=0:}\begin{equation*} \delta \int \frac{1}{2} \dot{s}^{2} d \tau=\delta \int \frac{1}{2}\left[\dot{t}^{2}-e^{2 t / R}\left(\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2}\right)\right] d \tau=0 \tag{3.1458} \end{equation*}(3.1458)δ12s˙2dτ=δ12[t˙2e2t/R(x˙2+y˙2+z˙2)]dτ=0
with the dot indicating differentiation with respect to the parameter τ τ tau\tauτ. The EulerLagrange equations for this variational problem are
(3.1459) t ¨ = 1 R e 2 t / R ( x ˙ 2 + y ˙ 2 + z ˙ 2 ) (3.1460) d d τ ( e 2 t / R x ˙ i ) = 0 (3.1459) t ¨ = 1 R e 2 t / R x ˙ 2 + y ˙ 2 + z ˙ 2 (3.1460) d d τ e 2 t / R x ˙ i = 0 {:[(3.1459)t^(¨)=-(1)/(R)e^(2t//R)(x^(˙)^(2)+y^(˙)^(2)+z^(˙)^(2))],[(3.1460)(d)/(d tau)(-e^(2t//R)x^(˙)^(i))=0]:}\begin{align*} \ddot{t} & =-\frac{1}{R} e^{2 t / R}\left(\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2}\right) \tag{3.1459}\\ \frac{d}{d \tau}\left(-e^{2 t / R} \dot{x}^{i}\right) & =0 \tag{3.1460} \end{align*}(3.1459)t¨=1Re2t/R(x˙2+y˙2+z˙2)(3.1460)ddτ(e2t/Rx˙i)=0
where i = 1 , 2 , 3 i = 1 , 2 , 3 i=1,2,3i=1,2,3i=1,2,3 and ( x 1 , x 2 , x 3 ) = ( x , y , z ) x 1 , x 2 , x 3 = ( x , y , z ) (x^(1),x^(2),x^(3))=(x,y,z)\left(x^{1}, x^{2}, x^{3}\right)=(x, y, z)(x1,x2,x3)=(x,y,z). The latter Euler-Lagrange equations can be integrated and give
(3.1461) x ˙ i = c i e 2 t / R (3.1461) x ˙ i = c i e 2 t / R {:(3.1461)x^(˙)^(i)=c^(i)e^(-2t//R):}\begin{equation*} \dot{x}^{i}=c^{i} e^{-2 t / R} \tag{3.1461} \end{equation*}(3.1461)x˙i=cie2t/R
for certain integration constants c i c i c^(i)c^{i}ci, which are fixed by the initial conditions. Thus, we find
(3.1462) d x i d x j = c i c j = const. (3.1462) d x i d x j = c i c j =  const.  {:(3.1462)(dx^(i))/(dx^(j))=(c^(i))/(c^(j))=" const. ":}\begin{equation*} \frac{d x^{i}}{d x^{j}}=\frac{c^{i}}{c^{j}}=\text { const. } \tag{3.1462} \end{equation*}(3.1462)dxidxj=cicj= const. 
for all i j i j i!=ji \neq jij, which can be solved in the following simple manner, e.g.,
(3.1463) x = a + c 1 z / c 3 , y = b + c 2 z / c 3 (3.1463) x = a + c 1 z / c 3 , y = b + c 2 z / c 3 {:(3.1463)x=a+c^(1)z//c^(3)","quad y=b+c^(2)z//c^(3):}\begin{equation*} x=a+c^{1} z / c^{3}, \quad y=b+c^{2} z / c^{3} \tag{3.1463} \end{equation*}(3.1463)x=a+c1z/c3,y=b+c2z/c3
with further integration constants a a aaa and b b bbb. This proves that all geodesics for the given metric are straight lines.
b) The photon moves on a lightlike trajectory along the x x xxx-axis, i.e., d s = d y = d s = d y = ds=dy=d s=d y=ds=dy= d z = 0 d z = 0 dz=0d z=0dz=0. This gives
(3.1464) d t = e t / R d x (3.1464) d t = e t / R d x {:(3.1464)dt=-e^(t//R)dx:}\begin{equation*} d t=-e^{t / R} d x \tag{3.1464} \end{equation*}(3.1464)dt=et/Rdx
where we assume the minus sign, since we want t ˙ > 0 t ˙ > 0 t^(˙) > 0\dot{t}>0t˙>0 and x ˙ < 0 x ˙ < 0 x^(˙) < 0\dot{x}<0x˙<0 at t = 0 t = 0 t=0t=0t=0. This implies
(3.1465) 0 t ( x ) e t / R d t = X 0 d x (3.1465) 0 t ( x ) e t / R d t = X 0 d x {:(3.1465)int_(0)^(t(x))e^(-t//R)dt=-int_(X)^(0)dx:}\begin{equation*} \int_{0}^{t(x)} e^{-t / R} d t=-\int_{X}^{0} d x \tag{3.1465} \end{equation*}(3.1465)0t(x)et/Rdt=X0dx
which yields
(3.1466) R ( e t / R 1 ) = X (3.1466) R e t / R 1 = X {:(3.1466)-R(e^(-t//R)-1)=X:}\begin{equation*} -R\left(e^{-t / R}-1\right)=X \tag{3.1466} \end{equation*}(3.1466)R(et/R1)=X
and thus, we obtain
(3.1467) t = R log ( 1 X / R ) (3.1467) t = R log ( 1 X / R ) {:(3.1467)t=-R log(1-X//R):}\begin{equation*} t=-R \log (1-X / R) \tag{3.1467} \end{equation*}(3.1467)t=Rlog(1X/R)

2.84

a) A suitable coordinate system for the problem is given by using spherical coordinates in the subspace spanned by the spatial directions of the five-dimensional Minkowski space, i.e.,
(3.1468) x 1 = r sin ( χ ) sin ( θ ) cos ( ϕ ) (3.1469) x 2 = r sin ( χ ) sin ( θ ) sin ( ϕ ) (3.1470) x 3 = r sin ( χ ) cos ( θ ) (3.1471) x 4 = r cos ( χ ) (3.1468) x 1 = r sin ( χ ) sin ( θ ) cos ( ϕ ) (3.1469) x 2 = r sin ( χ ) sin ( θ ) sin ( ϕ ) (3.1470) x 3 = r sin ( χ ) cos ( θ ) (3.1471) x 4 = r cos ( χ ) {:[(3.1468)x^(1)=r sin(chi)sin(theta)cos(phi)],[(3.1469)x^(2)=r sin(chi)sin(theta)sin(phi)],[(3.1470)x^(3)=r sin(chi)cos(theta)],[(3.1471)x^(4)=r cos(chi)]:}\begin{align*} x^{1} & =r \sin (\chi) \sin (\theta) \cos (\phi) \tag{3.1468}\\ x^{2} & =r \sin (\chi) \sin (\theta) \sin (\phi) \tag{3.1469}\\ x^{3} & =r \sin (\chi) \cos (\theta) \tag{3.1470}\\ x^{4} & =r \cos (\chi) \tag{3.1471} \end{align*}(3.1468)x1=rsin(χ)sin(θ)cos(ϕ)(3.1469)x2=rsin(χ)sin(θ)sin(ϕ)(3.1470)x3=rsin(χ)cos(θ)(3.1471)x4=rcos(χ)
In these coordinates, dS 4 dS 4 dS_(4)\mathrm{dS}_{4}dS4 is the restriction of the five-dimensional Minkowski metric to the hyperboloid t 2 r 2 = T 0 2 t 2 r 2 = T 0 2 t^(2)-r^(2)=-T_(0)^(2)t^{2}-r^{2}=-T_{0}^{2}t2r2=T02. This condition can be parametrized as
(3.1472) r = T 0 cosh ( τ / T 0 ) , t = T 0 sinh ( τ / T 0 ) (3.1472) r = T 0 cosh τ / T 0 , t = T 0 sinh τ / T 0 {:(3.1472)r=T_(0)cosh(tau//T_(0))","quad t=T_(0)sinh(tau//T_(0)):}\begin{equation*} r=T_{0} \cosh \left(\tau / T_{0}\right), \quad t=T_{0} \sinh \left(\tau / T_{0}\right) \tag{3.1472} \end{equation*}(3.1472)r=T0cosh(τ/T0),t=T0sinh(τ/T0)
In the coordinates ( τ , χ , θ , ϕ ) ( τ , χ , θ , ϕ ) (tau,chi,theta,phi)(\tau, \chi, \theta, \phi)(τ,χ,θ,ϕ), the line element becomes
(3.1473) d s 2 = d τ 2 T 0 2 cosh 2 ( τ / T 0 ) d Ω 2 (3.1473) d s 2 = d τ 2 T 0 2 cosh 2 τ / T 0 d Ω 2 {:(3.1473)ds^(2)=dtau^(2)-T_(0)^(2)cosh^(2)(tau//T_(0))dOmega^(2):}\begin{equation*} d s^{2}=d \tau^{2}-T_{0}^{2} \cosh ^{2}\left(\tau / T_{0}\right) d \Omega^{2} \tag{3.1473} \end{equation*}(3.1473)ds2=dτ2T02cosh2(τ/T0)dΩ2
where d Ω 2 = d χ 2 + sin 2 ( χ ) [ d θ 2 + sin 2 ( θ ) d ϕ 2 ] d Ω 2 = d χ 2 + sin 2 ( χ ) d θ 2 + sin 2 ( θ ) d ϕ 2 dOmega^(2)=dchi^(2)+sin^(2)(chi)[dtheta^(2)+sin^(2)(theta)dphi^(2)]d \Omega^{2}=d \chi^{2}+\sin ^{2}(\chi)\left[d \theta^{2}+\sin ^{2}(\theta) d \phi^{2}\right]dΩ2=dχ2+sin2(χ)[dθ2+sin2(θ)dϕ2] is the standard line element on the three-dimensional sphere.
b) The geodesic equations are given by varying the integral S = L d s S = L d s S=intLdsS=\int \mathcal{L} d sS=Lds, where
(3.1474) L = g i j x ˙ i x ˙ j = τ ˙ 2 T 0 2 cosh 2 ( τ / T 0 ) { χ ˙ 2 + sin 2 ( χ ) [ θ ˙ 2 + sin 2 ( θ ) ϕ ˙ 2 ] } (3.1474) L = g i j x ˙ i x ˙ j = τ ˙ 2 T 0 2 cosh 2 τ / T 0 χ ˙ 2 + sin 2 ( χ ) θ ˙ 2 + sin 2 ( θ ) ϕ ˙ 2 {:(3.1474)L=g_(ij)x^(˙)^(i)x^(˙)^(j)=tau^(˙)^(2)-T_(0)^(2)cosh^(2)(tau//T_(0)){chi^(˙)^(2)+sin^(2)(chi)[theta^(˙)^(2)+sin^(2)(theta)phi^(˙)^(2)]}:}\begin{equation*} \mathcal{L}=g_{i j} \dot{x}^{i} \dot{x}^{j}=\dot{\tau}^{2}-T_{0}^{2} \cosh ^{2}\left(\tau / T_{0}\right)\left\{\dot{\chi}^{2}+\sin ^{2}(\chi)\left[\dot{\theta}^{2}+\sin ^{2}(\theta) \dot{\phi}^{2}\right]\right\} \tag{3.1474} \end{equation*}(3.1474)L=gijx˙ix˙j=τ˙2T02cosh2(τ/T0){χ˙2+sin2(χ)[θ˙2+sin2(θ)ϕ˙2]}
From the Euler-Lagrange equation, we deduce
(3.1475) τ ¨ + T 0 2 sinh ( 2 τ / T 0 ) { χ ˙ 2 + sin 2 ( χ ) [ θ ˙ 2 + sin 2 ( θ ) ϕ ˙ 2 ] } = 0 (3.1475) τ ¨ + T 0 2 sinh 2 τ / T 0 χ ˙ 2 + sin 2 ( χ ) θ ˙ 2 + sin 2 ( θ ) ϕ ˙ 2 = 0 {:(3.1475)tau^(¨)+(T_(0))/(2)sinh(2tau//T_(0)){chi^(˙)^(2)+sin^(2)(chi)[theta^(˙)^(2)+sin^(2)(theta)phi^(˙)^(2)]}=0:}\begin{equation*} \ddot{\tau}+\frac{T_{0}}{2} \sinh \left(2 \tau / T_{0}\right)\left\{\dot{\chi}^{2}+\sin ^{2}(\chi)\left[\dot{\theta}^{2}+\sin ^{2}(\theta) \dot{\phi}^{2}\right]\right\}=0 \tag{3.1475} \end{equation*}(3.1475)τ¨+T02sinh(2τ/T0){χ˙2+sin2(χ)[θ˙2+sin2(θ)ϕ˙2]}=0
(3.1476) χ ¨ 1 2 sin ( 2 χ ) [ θ ˙ 2 + sin 2 ( θ ) ϕ ˙ 2 ] + 2 T 0 tanh ( τ / T 0 ) τ ˙ χ ˙ = 0 (3.1477) θ ¨ + 2 cot ( χ ) χ ˙ θ ˙ 1 2 sin ( 2 θ ) ϕ ˙ 2 + 2 T 0 tanh ( τ / T 0 ) τ ˙ θ ˙ = 0 (3.1478) ϕ ¨ + 2 cot ( χ ) χ ˙ ϕ ˙ + 2 cot ( θ ) θ ˙ ϕ ˙ + 2 T 0 tanh ( τ / T 0 ) τ ˙ ϕ ˙ = 0 (3.1476) χ ¨ 1 2 sin ( 2 χ ) θ ˙ 2 + sin 2 ( θ ) ϕ ˙ 2 + 2 T 0 tanh τ / T 0 τ ˙ χ ˙ = 0 (3.1477) θ ¨ + 2 cot ( χ ) χ ˙ θ ˙ 1 2 sin ( 2 θ ) ϕ ˙ 2 + 2 T 0 tanh τ / T 0 τ ˙ θ ˙ = 0 (3.1478) ϕ ¨ + 2 cot ( χ ) χ ˙ ϕ ˙ + 2 cot ( θ ) θ ˙ ϕ ˙ + 2 T 0 tanh τ / T 0 τ ˙ ϕ ˙ = 0 {:[(3.1476)chi^(¨)-(1)/(2)sin(2chi)[theta^(˙)^(2)+sin^(2)(theta)phi^(˙)^(2)]+(2)/(T_(0))tanh(tau//T_(0))tau^(˙)chi^(˙)=0],[(3.1477)theta^(¨)+2cot(chi)chi^(˙)theta^(˙)-(1)/(2)sin(2theta)phi^(˙)^(2)+(2)/(T_(0))tanh(tau//T_(0))tau^(˙)theta^(˙)=0],[(3.1478)phi^(¨)+2cot(chi)chi^(˙)phi^(˙)+2cot(theta)theta^(˙)phi^(˙)+(2)/(T_(0))tanh(tau//T_(0))tau^(˙)phi^(˙)=0]:}\begin{align*} \ddot{\chi}-\frac{1}{2} \sin (2 \chi)\left[\dot{\theta}^{2}+\sin ^{2}(\theta) \dot{\phi}^{2}\right]+\frac{2}{T_{0}} \tanh \left(\tau / T_{0}\right) \dot{\tau} \dot{\chi} & =0 \tag{3.1476}\\ \ddot{\theta}+2 \cot (\chi) \dot{\chi} \dot{\theta}-\frac{1}{2} \sin (2 \theta) \dot{\phi}^{2}+\frac{2}{T_{0}} \tanh \left(\tau / T_{0}\right) \dot{\tau} \dot{\theta} & =0 \tag{3.1477}\\ \ddot{\phi}+2 \cot (\chi) \dot{\chi} \dot{\phi}+2 \cot (\theta) \dot{\theta} \dot{\phi}+\frac{2}{T_{0}} \tanh \left(\tau / T_{0}\right) \dot{\tau} \dot{\phi} & =0 \tag{3.1478} \end{align*}(3.1476)χ¨12sin(2χ)[θ˙2+sin2(θ)ϕ˙2]+2T0tanh(τ/T0)τ˙χ˙=0(3.1477)θ¨+2cot(χ)χ˙θ˙12sin(2θ)ϕ˙2+2T0tanh(τ/T0)τ˙θ˙=0(3.1478)ϕ¨+2cot(χ)χ˙ϕ˙+2cot(θ)θ˙ϕ˙+2T0tanh(τ/T0)τ˙ϕ˙=0
c) Comparing with the Robertson-Walker metric, we obtain
(3.1479) S ( τ ) = T 0 cosh ( τ / T 0 ) (3.1479) S ( τ ) = T 0 cosh τ / T 0 {:(3.1479)S(tau)=T_(0)cosh(tau//T_(0)):}\begin{equation*} S(\tau)=T_{0} \cosh \left(\tau / T_{0}\right) \tag{3.1479} \end{equation*}(3.1479)S(τ)=T0cosh(τ/T0)
i.e., a universe which is contracting for τ < 0 τ < 0 tau < 0\tau<0τ<0 and expanding for τ > 0 τ > 0 tau > 0\tau>0τ>0.
If we compute the Ricci tensor and Ricci scalar for dS 4 dS 4 dS_(4)\mathrm{dS}_{4}dS4, then we obtain
(3.1480) R μ ν = 3 T 0 2 g μ ν , R = 12 T 0 2 (3.1480) R μ ν = 3 T 0 2 g μ ν , R = 12 T 0 2 {:(3.1480)R_(mu nu)=(3)/(T_(0)^(2))g_(mu nu)","quad R=(12)/(T_(0)^(2)):}\begin{equation*} R_{\mu \nu}=\frac{3}{T_{0}^{2}} g_{\mu \nu}, \quad R=\frac{12}{T_{0}^{2}} \tag{3.1480} \end{equation*}(3.1480)Rμν=3T02gμν,R=12T02
Thus, the Einstein tensor is given by
(3.1481) G μ ν = R μ ν 1 2 g μ ν R = 3 T 0 2 g μ ν = 8 π G T μ ν (3.1481) G μ ν = R μ ν 1 2 g μ ν R = 3 T 0 2 g μ ν = 8 π G T μ ν {:(3.1481)G_(mu nu)=R_(mu nu)-(1)/(2)g_(mu nu)R=-(3)/(T_(0)^(2))g_(mu nu)=8pi GT_(mu nu):}\begin{equation*} G_{\mu \nu}=R_{\mu \nu}-\frac{1}{2} g_{\mu \nu} R=-\frac{3}{T_{0}^{2}} g_{\mu \nu}=8 \pi G T_{\mu \nu} \tag{3.1481} \end{equation*}(3.1481)Gμν=Rμν12gμνR=3T02gμν=8πGTμν
This gives T μ ν = Λ g μ ν / ( 8 π G ) T μ ν = Λ g μ ν / ( 8 π G ) T_(mu nu)=-Lambdag_(mu nu)//(8pi G)T_{\mu \nu}=-\Lambda g_{\mu \nu} /(8 \pi G)Tμν=Λgμν/(8πG), which is typical for so-called dark energy ( Λ Λ Lambda\LambdaΛ is also know as cosmological constant, in dS 4 dS 4 dS_(4)\mathrm{dS}_{4}dS4 we have a positive cosmological constant Λ = 3 / T 0 2 Λ = 3 / T 0 2 Lambda=3//T_(0)^(2)\Lambda=3 / T_{0}^{2}Λ=3/T02 ).

2.85

a) Let the Lagrangian be
(3.1482) L = 1 2 [ ( r 2 1 ) t ˙ 2 1 r 2 1 r ˙ 2 ] , (3.1482) L = 1 2 r 2 1 t ˙ 2 1 r 2 1 r ˙ 2 , {:(3.1482)L=(1)/(2)[(r^(2)-1)t^(˙)^(2)-(1)/(r^(2)-1)r^(˙)^(2)]",":}\begin{equation*} \mathcal{L}=\frac{1}{2}\left[\left(r^{2}-1\right) \dot{t}^{2}-\frac{1}{r^{2}-1} \dot{r}^{2}\right], \tag{3.1482} \end{equation*}(3.1482)L=12[(r21)t˙21r21r˙2],
see Problem 2.27. The trajectory of a light ray then obeys L = 0 L = 0 L=0\mathcal{L}=0L=0, i.e., ( r 2 1 ) 2 t ˙ 2 = r 2 1 2 t ˙ 2 = (r^(2)-1)^(2)t^(˙)^(2)=\left(r^{2}-1\right)^{2} \dot{t}^{2}=(r21)2t˙2= r ˙ 2 r ˙ 2 r^(˙)^(2)\dot{r}^{2}r˙2. Taking the square root, we can solve this equation by separation and find that
(3.1483) d r r 2 1 = 1 2 ( 1 r 1 1 r + 1 ) = ± d t (3.1483) d r r 2 1 = 1 2 1 r 1 1 r + 1 = ± d t {:(3.1483)(dr)/(r^(2)-1)=(1)/(2)((1)/(r-1)-(1)/(r+1))=+-dt:}\begin{equation*} \frac{d r}{r^{2}-1}=\frac{1}{2}\left(\frac{1}{r-1}-\frac{1}{r+1}\right)= \pm d t \tag{3.1483} \end{equation*}(3.1483)drr21=12(1r11r+1)=±dt
Integrating, we obtain
(3.1484) 1 2 ( ln r 1 r 0 1 ln r + 1 r 0 + 1 ) = ± t (3.1484) 1 2 ln r 1 r 0 1 ln r + 1 r 0 + 1 = ± t {:(3.1484)(1)/(2)(ln((r-1)/(r_(0)-1))-ln((r+1)/(r_(0)+1)))=+-t:}\begin{equation*} \frac{1}{2}\left(\ln \frac{r-1}{r_{0}-1}-\ln \frac{r+1}{r_{0}+1}\right)= \pm t \tag{3.1484} \end{equation*}(3.1484)12(lnr1r01lnr+1r0+1)=±t
where we integrated the left-hand side from r 0 r 0 r_(0)r_{0}r0 to r = r ( t ) r = r ( t ) r=r(t)r=r(t)r=r(t) and the right-hand side from t 0 t 0 t_(0)t_{0}t0 to t t ttt so that r ( 0 ) = r 0 r ( 0 ) = r 0 r(0)=r_(0)r(0)=r_{0}r(0)=r0. Thus, we can write this as
(3.1485) r 1 r + 1 = r 0 1 r 0 + 1 exp ( ± 2 t ) (3.1485) r 1 r + 1 = r 0 1 r 0 + 1 exp ( ± 2 t ) {:(3.1485)(r-1)/(r+1)=(r_(0)-1)/(r_(0)+1)exp(+-2t):}\begin{equation*} \frac{r-1}{r+1}=\frac{r_{0}-1}{r_{0}+1} \exp ( \pm 2 t) \tag{3.1485} \end{equation*}(3.1485)r1r+1=r01r0+1exp(±2t)
and after some computations, we find
(3.1486) r ( t ) = r 0 + 1 + ( r 0 1 ) exp ( ± 2 t ) r 0 + 1 ( r 0 1 ) exp ( ± 2 t ) (3.1486) r ( t ) = r 0 + 1 + r 0 1 exp ( ± 2 t ) r 0 + 1 r 0 1 exp ( ± 2 t ) {:(3.1486)r(t)=(r_(0)+1+(r_(0)-1)exp(+-2t))/(r_(0)+1-(r_(0)-1)exp(+-2t)):}\begin{equation*} r(t)=\frac{r_{0}+1+\left(r_{0}-1\right) \exp ( \pm 2 t)}{r_{0}+1-\left(r_{0}-1\right) \exp ( \pm 2 t)} \tag{3.1486} \end{equation*}(3.1486)r(t)=r0+1+(r01)exp(±2t)r0+1(r01)exp(±2t)
Note that the different signs ± ± +-\pm± correspond to the two possible directions of the light ray.
b) The trajectory of a particle can also be obtained from the Lagrangian L L L\mathcal{L}L. Using L / t = 0 L / t = 0 delL//del t=0\partial \mathcal{L} / \partial t=0L/t=0, we find a conservation law
(3.1487) d d τ L t ˙ = 0 (3.1487) d d τ L t ˙ = 0 {:(3.1487)(d)/(d tau)*(delL)/(del(t^(˙)))=0:}\begin{equation*} \frac{d}{d \tau} \cdot \frac{\partial \mathcal{L}}{\partial \dot{t}}=0 \tag{3.1487} \end{equation*}(3.1487)ddτLt˙=0
i.e.,
(3.1488) L t ˙ = ( r 2 1 ) t ˙ = c 1 (3.1488) L t ˙ = r 2 1 t ˙ = c 1 {:(3.1488)(delL)/(del(t^(˙)))=(r^(2)-1)t^(˙)=c_(1):}\begin{equation*} \frac{\partial \mathcal{L}}{\partial \dot{t}}=\left(r^{2}-1\right) \dot{t}=c_{1} \tag{3.1488} \end{equation*}(3.1488)Lt˙=(r21)t˙=c1
for some constant c 1 c 1 c_(1)c_{1}c1. To compute c 1 c 1 c_(1)c_{1}c1, we set t = 0 t = 0 t=0t=0t=0 for which it holds that r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0 and i = γ = 1 i = γ = 1 i=gamma=1i=\gamma=1i=γ=1, since the velocity is zero. Thus, we find c 1 = r 0 2 1 c 1 = r 0 2 1 c_(1)=r_(0)^(2)-1c_{1}=r_{0}^{2}-1c1=r021, and therefore, we have
(3.1489) t ˙ = r 0 2 1 r 2 1 (3.1489) t ˙ = r 0 2 1 r 2 1 {:(3.1489)t^(˙)=(r_(0)^(2)-1)/(r^(2)-1):}\begin{equation*} \dot{t}=\frac{r_{0}^{2}-1}{r^{2}-1} \tag{3.1489} \end{equation*}(3.1489)t˙=r021r21
A second conservation law is L = c 2 L = c 2 L=c_(2)\mathcal{L}=c_{2}L=c2 for some other constant c 2 c 2 c_(2)c_{2}c2. Inserting again t = 0 t = 0 t=0t=0t=0, we find c 2 = ( r 0 2 1 ) / 2 c 2 = r 0 2 1 / 2 c_(2)=(r_(0)^(2)-1)//2c_{2}=\left(r_{0}^{2}-1\right) / 2c2=(r021)/2, since t ˙ = 1 , r ˙ = 0 t ˙ = 1 , r ˙ = 0 t^(˙)=1,r^(˙)=0\dot{t}=1, \dot{r}=0t˙=1,r˙=0, and r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0 at t = 0 t = 0 t=0t=0t=0. Combining the first and the second conservation laws and multiplying with a factor of two, we obtain
( r 2 1 ) t ˙ 2 1 r 2 1 r ˙ 2 = ( r 0 2 1 ) 2 r ˙ 2 r 2 1 = r 0 2 1 r ˙ 2 = ( r 0 2 1 ) ( r 0 2 r 2 ) r 2 1 t ˙ 2 1 r 2 1 r ˙ 2 = r 0 2 1 2 r ˙ 2 r 2 1 = r 0 2 1 r ˙ 2 = r 0 2 1 r 0 2 r 2 (r^(2)-1)t^(˙)^(2)-(1)/(r^(2)-1)r^(˙)^(2)=((r_(0)^(2)-1)^(2)-r^(˙)^(2))/(r^(2)-1)=r_(0)^(2)-1quad=>quadr^(˙)^(2)=(r_(0)^(2)-1)(r_(0)^(2)-r^(2))\left(r^{2}-1\right) \dot{t}^{2}-\frac{1}{r^{2}-1} \dot{r}^{2}=\frac{\left(r_{0}^{2}-1\right)^{2}-\dot{r}^{2}}{r^{2}-1}=r_{0}^{2}-1 \quad \Rightarrow \quad \dot{r}^{2}=\left(r_{0}^{2}-1\right)\left(r_{0}^{2}-r^{2}\right)(r21)t˙21r21r˙2=(r021)2r˙2r21=r021r˙2=(r021)(r02r2).
Separating this equation, we obtain
(3.1491) d τ = d r ( r 0 2 1 ) ( r 0 2 r 2 ) (3.1491) d τ = d r r 0 2 1 r 0 2 r 2 {:(3.1491)d tau=-(dr)/(sqrt((r_(0)^(2)-1)(r_(0)^(2)-r^(2)))):}\begin{equation*} d \tau=-\frac{d r}{\sqrt{\left(r_{0}^{2}-1\right)\left(r_{0}^{2}-r^{2}\right)}} \tag{3.1491} \end{equation*}(3.1491)dτ=dr(r021)(r02r2)
where we choose the sign of the square root so that r ˙ < 0 r ˙ < 0 r^(˙) < 0\dot{r}<0r˙<0. Integrating the left-hand side from 0 to τ τ tau\tauτ and the right-hand side from r 0 r 0 r_(0)r_{0}r0 to r 1 r 1 r_(1)r_{1}r1 yields the proper time
(3.1492) τ = r 0 r 1 d r ( r 0 2 1 ) ( r 0 2 r 2 ) = 1 r 0 2 1 ( π 2 arctan r 1 r 0 2 r 1 2 ) (3.1492) τ = r 0 r 1 d r r 0 2 1 r 0 2 r 2 = 1 r 0 2 1 π 2 arctan r 1 r 0 2 r 1 2 {:(3.1492)tau=-int_(r_(0))^(r_(1))(dr)/(sqrt((r_(0)^(2)-1)(r_(0)^(2)-r^(2))))=(1)/(sqrt(r_(0)^(2)-1))((pi)/(2)-arctan((r_(1))/(sqrt(r_(0)^(2)-r_(1)^(2))))):}\begin{equation*} \tau=-\int_{r_{0}}^{r_{1}} \frac{d r}{\sqrt{\left(r_{0}^{2}-1\right)\left(r_{0}^{2}-r^{2}\right)}}=\frac{1}{\sqrt{r_{0}^{2}-1}}\left(\frac{\pi}{2}-\arctan \frac{r_{1}}{\sqrt{r_{0}^{2}-r_{1}^{2}}}\right) \tag{3.1492} \end{equation*}(3.1492)τ=r0r1dr(r021)(r02r2)=1r021(π2arctanr1r02r12)
To compute the coordinate time, we use the first conservation law and the result for d τ d τ d taud \taudτ, i.e.,
(3.1493) d t = r 0 2 1 r 2 1 d τ = r 0 2 1 r 2 1 d r ( r 0 2 1 ) ( r 0 2 r 2 ) = r 0 2 1 d r ( r 2 1 ) r 0 2 r 2 (3.1493) d t = r 0 2 1 r 2 1 d τ = r 0 2 1 r 2 1 d r r 0 2 1 r 0 2 r 2 = r 0 2 1 d r r 2 1 r 0 2 r 2 {:(3.1493)dt=(r_(0)^(2)-1)/(r^(2)-1)d tau=-(r_(0)^(2)-1)/(r^(2)-1)(dr)/(sqrt((r_(0)^(2)-1)(r_(0)^(2)-r^(2))))=-sqrt(r_(0)^(2)-1)(dr)/((r^(2)-1)sqrt(r_(0)^(2)-r^(2))):}\begin{equation*} d t=\frac{r_{0}^{2}-1}{r^{2}-1} d \tau=-\frac{r_{0}^{2}-1}{r^{2}-1} \frac{d r}{\sqrt{\left(r_{0}^{2}-1\right)\left(r_{0}^{2}-r^{2}\right)}}=-\sqrt{r_{0}^{2}-1} \frac{d r}{\left(r^{2}-1\right) \sqrt{r_{0}^{2}-r^{2}}} \tag{3.1493} \end{equation*}(3.1493)dt=r021r21dτ=r021r21dr(r021)(r02r2)=r021dr(r21)r02r2
Thus, integrating, we have
(3.1494) t = r 0 2 1 r 0 r 1 d r ( r 2 1 ) r 0 2 r 2 (3.1494) t = r 0 2 1 r 0 r 1 d r r 2 1 r 0 2 r 2 {:(3.1494)t=-sqrt(r_(0)^(2)-1)int_(r_(0))^(r_(1))(dr)/((r^(2)-1)sqrt(r_(0)^(2)-r^(2))):}\begin{equation*} t=-\sqrt{r_{0}^{2}-1} \int_{r_{0}}^{r_{1}} \frac{d r}{\left(r^{2}-1\right) \sqrt{r_{0}^{2}-r^{2}}} \tag{3.1494} \end{equation*}(3.1494)t=r021r0r1dr(r21)r02r2
When r 1 1 r 1 1 r_(1)rarr1r_{1} \rightarrow 1r11, the integral diverges as t ln ( r 1 1 ) / 2 t ln r 1 1 / 2 t∼-ln(r_(1)-1)//2t \sim-\ln \left(r_{1}-1\right) / 2tln(r11)/2. The point r = 1 r = 1 r=1r=1r=1 is a singularity similar to the singularity of the Schwarzschild metric at the event horizon: Our computation above shows that the proper time for a particle falling toward r = 1 r = 1 r=1r=1r=1 is finite, whereas the coordinate time for that is infinite. Thus, this suggests that r = 1 r = 1 r=1r=1r=1 is a coordinate singularity.

2.86

a) Using the given coordinate transformations, we have
(3.1495) d U = cos ( α ) d t t sin ( α ) d α (3.1496) d V = sin ( α ) d t + t cos ( α ) d α (3.1497) d X = sin ( θ ) cos ( φ ) d r + r cos ( θ ) cos ( φ ) d θ r sin ( θ ) sin ( φ ) d φ (3.1498) d Y = sin ( θ ) sin ( φ ) d r + r cos ( θ ) sin ( φ ) d θ + r sin ( θ ) cos ( φ ) d φ (3.1499) d Z = cos ( θ ) d r r sin ( θ ) d θ (3.1495) d U = cos ( α ) d t t sin ( α ) d α (3.1496) d V = sin ( α ) d t + t cos ( α ) d α (3.1497) d X = sin ( θ ) cos ( φ ) d r + r cos ( θ ) cos ( φ ) d θ r sin ( θ ) sin ( φ ) d φ (3.1498) d Y = sin ( θ ) sin ( φ ) d r + r cos ( θ ) sin ( φ ) d θ + r sin ( θ ) cos ( φ ) d φ (3.1499) d Z = cos ( θ ) d r r sin ( θ ) d θ {:[(3.1495)dU=cos(alpha)dt-t sin(alpha)d alpha],[(3.1496)dV=sin(alpha)dt+t cos(alpha)d alpha],[(3.1497)dX=sin(theta)cos(varphi)dr+r cos(theta)cos(varphi)d theta-r sin(theta)sin(varphi)d varphi],[(3.1498)dY=sin(theta)sin(varphi)dr+r cos(theta)sin(varphi)d theta+r sin(theta)cos(varphi)d varphi],[(3.1499)dZ=cos(theta)dr-r sin(theta)d theta]:}\begin{align*} d U & =\cos (\alpha) d t-t \sin (\alpha) d \alpha \tag{3.1495}\\ d V & =\sin (\alpha) d t+t \cos (\alpha) d \alpha \tag{3.1496}\\ d X & =\sin (\theta) \cos (\varphi) d r+r \cos (\theta) \cos (\varphi) d \theta-r \sin (\theta) \sin (\varphi) d \varphi \tag{3.1497}\\ d Y & =\sin (\theta) \sin (\varphi) d r+r \cos (\theta) \sin (\varphi) d \theta+r \sin (\theta) \cos (\varphi) d \varphi \tag{3.1498}\\ d Z & =\cos (\theta) d r-r \sin (\theta) d \theta \tag{3.1499} \end{align*}(3.1495)dU=cos(α)dttsin(α)dα(3.1496)dV=sin(α)dt+tcos(α)dα(3.1497)dX=sin(θ)cos(φ)dr+rcos(θ)cos(φ)dθrsin(θ)sin(φ)dφ(3.1498)dY=sin(θ)sin(φ)dr+rcos(θ)sin(φ)dθ+rsin(θ)cos(φ)dφ(3.1499)dZ=cos(θ)drrsin(θ)dθ
Thus, the metric of AdS 4 AdS 4 AdS_(4)\mathrm{AdS}_{4}AdS4 in the coordinates α , λ , θ α , λ , θ alpha,lambda,theta\alpha, \lambda, \thetaα,λ,θ, and φ φ varphi\varphiφ is given by
(3.1500) d s 2 = d t 2 + t 2 d α 2 d r 2 r 2 d Ω 2 = cosh ( λ ) 2 d α 2 d λ 2 sinh ( λ ) 2 d Ω 2 (3.1500) d s 2 = d t 2 + t 2 d α 2 d r 2 r 2 d Ω 2 = cosh ( λ ) 2 d α 2 d λ 2 sinh ( λ ) 2 d Ω 2 {:(3.1500)ds^(2)=dt^(2)+t^(2)dalpha^(2)-dr^(2)-r^(2)dOmega^(2)=cosh(lambda)^(2)dalpha^(2)-dlambda^(2)-sinh(lambda)^(2)dOmega^(2):}\begin{equation*} d s^{2}=d t^{2}+t^{2} d \alpha^{2}-d r^{2}-r^{2} d \Omega^{2}=\cosh (\lambda)^{2} d \alpha^{2}-d \lambda^{2}-\sinh (\lambda)^{2} d \Omega^{2} \tag{3.1500} \end{equation*}(3.1500)ds2=dt2+t2dα2dr2r2dΩ2=cosh(λ)2dα2dλ2sinh(λ)2dΩ2
where d Ω 2 = d θ 2 + sin ( θ ) 2 d φ 2 d Ω 2 = d θ 2 + sin ( θ ) 2 d φ 2 dOmega^(2)=dtheta^(2)+sin(theta)^(2)dvarphi^(2)d \Omega^{2}=d \theta^{2}+\sin (\theta)^{2} d \varphi^{2}dΩ2=dθ2+sin(θ)2dφ2.
b) To find the trajectories, we can solve (since θ ˙ = φ ˙ = 0 θ ˙ = φ ˙ = 0 theta^(˙)=varphi^(˙)=0\dot{\theta}=\dot{\varphi}=0θ˙=φ˙=0 )
(3.1501) L = 1 2 [ cosh ( λ ) 2 α ˙ 2 λ ˙ 2 ] = 0 (3.1501) L = 1 2 cosh ( λ ) 2 α ˙ 2 λ ˙ 2 = 0 {:(3.1501)L=(1)/(2)[cosh(lambda)^(2)alpha^(˙)^(2)-lambda^(˙)^(2)]=0:}\begin{equation*} \mathcal{L}=\frac{1}{2}\left[\cosh (\lambda)^{2} \dot{\alpha}^{2}-\dot{\lambda}^{2}\right]=0 \tag{3.1501} \end{equation*}(3.1501)L=12[cosh(λ)2α˙2λ˙2]=0
This gives
d λ cosh ( λ ) = ± d α 2 arctan tanh λ 2 = ± ( α α 0 ) (3.1502) λ = ± 2 artanh ( tan α α 0 2 ) d λ cosh ( λ ) = ± d α 2 arctan tanh λ 2 = ± α α 0 (3.1502) λ = ± 2 artanh tan α α 0 2 {:[(d lambda)/(cosh(lambda))=+-d alpha=>2arctan tanh((lambda)/(2))=+-(alpha-alpha_(0))],[(3.1502)=>lambda=+-2artanh(tan((alpha-alpha_(0))/(2)))]:}\begin{align*} \frac{d \lambda}{\cosh (\lambda)}= \pm d \alpha & \Rightarrow 2 \arctan \tanh \frac{\lambda}{2}= \pm\left(\alpha-\alpha_{0}\right) \\ & \Rightarrow \lambda= \pm 2 \operatorname{artanh}\left(\tan \frac{\alpha-\alpha_{0}}{2}\right) \tag{3.1502} \end{align*}dλcosh(λ)=±dα2arctantanhλ2=±(αα0)(3.1502)λ=±2artanh(tanαα02)
c) The Euler-Lagrange equations following from the variational problem defined in the hints are
(3.1503) X ¨ a = 2 λ X a , X a X a 1 = 0 (3.1503) X ¨ a = 2 λ X a , X a X a 1 = 0 {:(3.1503)X^(¨)^(a)=2lambdaX^(a)","quadX^(a)X_(a)-1=0:}\begin{equation*} \ddot{X}^{a}=2 \lambda X^{a}, \quad X^{a} X_{a}-1=0 \tag{3.1503} \end{equation*}(3.1503)X¨a=2λXa,XaXa1=0
Contracting the first equation X ¨ a = 2 λ X a X ¨ a = 2 λ X a X^(¨)^(a)=2lambdaX^(a)\ddot{X}^{a}=2 \lambda X^{a}X¨a=2λXa with X a X a X_(a)X_{a}Xa and using the second equation X a X a = 1 X a X a = 1 X^(a)X_(a)=1X^{a} X_{a}=1XaXa=1, we obtain
(3.1504) 2 λ = X ¨ a X a = d d τ ( X ˙ a X a ) X ˙ a X ˙ a (3.1504) 2 λ = X ¨ a X a = d d τ X ˙ a X a X ˙ a X ˙ a {:(3.1504)2lambda=X^(¨)^(a)X_(a)=(d)/(d tau)(X^(˙)^(a)X_(a))-X^(˙)^(a)X^(˙)_(a):}\begin{equation*} 2 \lambda=\ddot{X}^{a} X_{a}=\frac{d}{d \tau}\left(\dot{X}^{a} X_{a}\right)-\dot{X}^{a} \dot{X}_{a} \tag{3.1504} \end{equation*}(3.1504)2λ=X¨aXa=ddτ(X˙aXa)X˙aX˙a
but X ˙ a X a = 0 X ˙ a X a = 0 X^(˙)^(a)X_(a)=0\dot{X}^{a} X_{a}=0X˙aXa=0, which follows from differentiating X a X a = 1 X a X a = 1 X^(a)X_(a)=1X^{a} X_{a}=1XaXa=1 with respect to τ τ tau\tauτ, and thus, we have
(3.1505) 2 λ = X ˙ a X ˙ a (3.1505) 2 λ = X ˙ a X ˙ a {:(3.1505)2lambda=-X^(˙)^(a)X^(˙)_(a):}\begin{equation*} 2 \lambda=-\dot{X}^{a} \dot{X}_{a} \tag{3.1505} \end{equation*}(3.1505)2λ=X˙aX˙a
To prove that this is a constant of motion, as claimed in the hints, we compute
(3.1506) d d τ ( X ˙ a X ˙ a ) = 2 X ¨ a X ˙ a = 4 λ X a X ˙ a = 0 (3.1506) d d τ X ˙ a X ˙ a = 2 X ¨ a X ˙ a = 4 λ X a X ˙ a = 0 {:(3.1506)(d)/(d tau)(X^(˙)^(a)X^(˙)_(a))=2X^(¨)^(a)X^(˙)_(a)=4lambdaX^(a)X^(˙)_(a)=0:}\begin{equation*} \frac{d}{d \tau}\left(\dot{X}^{a} \dot{X}_{a}\right)=2 \ddot{X}^{a} \dot{X}_{a}=4 \lambda X^{a} \dot{X}_{a}=0 \tag{3.1506} \end{equation*}(3.1506)ddτ(X˙aX˙a)=2X¨aX˙a=4λXaX˙a=0
Thus, we can set X ˙ a X ˙ a = 0 X ˙ a X ˙ a = 0 X^(˙)^(a)X^(˙)_(a)=0\dot{X}^{a} \dot{X}_{a}=0X˙aX˙a=0, which implies that X ¨ a = 0 X ¨ a = 0 X^(¨)^(a)=0\ddot{X}^{a}=0X¨a=0.

2.87

a) We determine the Christoffel symbols by deriving the geodesic equation with L = g μ ν x ˙ μ x ˙ ν = t ˙ 2 a ( t ) 2 x ˙ 2 L = g μ ν x ˙ μ x ˙ ν = t ˙ 2 a ( t ) 2 x ˙ 2 L=g_(mu nu)x^(˙)^(mu)x^(˙)^(nu)=t^(˙)^(2)-a(t)^(2)x^(˙)^(2)\mathcal{L}=g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=\dot{t}^{2}-a(t)^{2} \dot{x}^{2}L=gμνx˙μx˙ν=t˙2a(t)2x˙2. We obtain
(3.1507) d d τ L t ˙ L t = 2 t ¨ + 2 a a x ˙ 2 = 0 t ¨ + a a x ˙ 2 = 0 (3.1508) d d τ L x ˙ L x = d d τ ( 2 a 2 x ˙ ) = 2 a 2 x ¨ + 4 a a t ˙ x ˙ = 0 x ¨ + 2 a a t ˙ x ˙ = 0 (3.1507) d d τ L t ˙ L t = 2 t ¨ + 2 a a x ˙ 2 = 0 t ¨ + a a x ˙ 2 = 0 (3.1508) d d τ L x ˙ L x = d d τ 2 a 2 x ˙ = 2 a 2 x ¨ + 4 a a t ˙ x ˙ = 0 x ¨ + 2 a a t ˙ x ˙ = 0 {:[(3.1507)(d)/(d tau)(delL)/(del(t^(˙)))-(delL)/(del t)=2t^(¨)+2aa^(')x^(˙)^(2)=0quad<=>quadt^(¨)+aa^(')x^(˙)^(2)=0],[(3.1508)(d)/(d tau)(delL)/(del(x^(˙)))-(delL)/(del x)=(d)/(d tau)(2a^(2)(x^(˙)))=2a^(2)x^(¨)+4aa^(')t^(˙)x^(˙)=0quad<=>quadx^(¨)+2(a^('))/(a)t^(˙)x^(˙)=0]:}\begin{align*} & \frac{d}{d \tau} \frac{\partial \mathcal{L}}{\partial \dot{t}}-\frac{\partial \mathcal{L}}{\partial t}=2 \ddot{t}+2 a a^{\prime} \dot{x}^{2}=0 \quad \Leftrightarrow \quad \ddot{t}+a a^{\prime} \dot{x}^{2}=0 \tag{3.1507}\\ & \frac{d}{d \tau} \frac{\partial \mathcal{L}}{\partial \dot{x}}-\frac{\partial \mathcal{L}}{\partial x}=\frac{d}{d \tau}\left(2 a^{2} \dot{x}\right)=2 a^{2} \ddot{x}+4 a a^{\prime} \dot{t} \dot{x}=0 \quad \Leftrightarrow \quad \ddot{x}+2 \frac{a^{\prime}}{a} \dot{t} \dot{x}=0 \tag{3.1508} \end{align*}(3.1507)ddτLt˙Lt=2t¨+2aax˙2=0t¨+aax˙2=0(3.1508)ddτLx˙Lx=ddτ(2a2x˙)=2a2x¨+4aat˙x˙=0x¨+2aat˙x˙=0
with a ( t ) = d a ( t ) / d t a ( t ) = d a ( t ) / d t a^(')(t)=da(t)//dta^{\prime}(t)=d a(t) / d ta(t)=da(t)/dt. Comparing with the general form of the geodesic equations x ¨ λ + Γ μ ν λ x ˙ μ x ˙ ν = 0 x ¨ λ + Γ μ ν λ x ˙ μ x ˙ ν = 0 x^(¨)^(lambda)+Gamma_(mu nu)^(lambda)x^(˙)^(mu)x^(˙)^(nu)=0\ddot{x}^{\lambda}+\Gamma_{\mu \nu}^{\lambda} \dot{x}^{\mu} \dot{x}^{\nu}=0x¨λ+Γμνλx˙μx˙ν=0, we can read off the following nonzero Christoffel symbols
(3.1509) Γ x x t = a a , Γ t x x = Γ x t x = a a (3.1509) Γ x x t = a a , Γ t x x = Γ x t x = a a {:(3.1509)Gamma_(xx)^(t)=aa^(')","quadGamma_(tx)^(x)=Gamma_(xt)^(x)=(a^('))/(a):}\begin{equation*} \Gamma_{x x}^{t}=a a^{\prime}, \quad \Gamma_{t x}^{x}=\Gamma_{x t}^{x}=\frac{a^{\prime}}{a} \tag{3.1509} \end{equation*}(3.1509)Γxxt=aa,Γtxx=Γxtx=aa
Now, we compute
R x t x t = t Γ x x t x Γ t x t + Γ t α t Γ x x α Γ x α t Γ t x α = t Γ x x t Γ x x t Γ t x x (3.1510) = t ( a a ) a a a a = ( a ) 2 + a a ( a ) 2 = a a R x t x t = t Γ x x t x Γ t x t + Γ t α t Γ x x α Γ x α t Γ t x α = t Γ x x t Γ x x t Γ t x x (3.1510) = t a a a a a a = a 2 + a a a 2 = a a {:[R_(xtx)^(t)=del_(t)Gamma_(xx)^(t)-del_(x)Gamma_(tx)^(t)+Gamma_(t alpha)^(t)Gamma_(xx)^(alpha)-Gamma_(x alpha)^(t)Gamma_(tx)^(alpha)=del_(t)Gamma_(xx)^(t)-Gamma_(xx)^(t)Gamma_(tx)^(x)],[(3.1510)=del_(t)(aa^('))-aa^(')(a^('))/(a)=(a^('))^(2)+aa^('')-(a^('))^(2)=aa^('')]:}\begin{align*} R_{x t x}^{t} & =\partial_{t} \Gamma_{x x}^{t}-\partial_{x} \Gamma_{t x}^{t}+\Gamma_{t \alpha}^{t} \Gamma_{x x}^{\alpha}-\Gamma_{x \alpha}^{t} \Gamma_{t x}^{\alpha}=\partial_{t} \Gamma_{x x}^{t}-\Gamma_{x x}^{t} \Gamma_{t x}^{x} \\ & =\partial_{t}\left(a a^{\prime}\right)-a a^{\prime} \frac{a^{\prime}}{a}=\left(a^{\prime}\right)^{2}+a a^{\prime \prime}-\left(a^{\prime}\right)^{2}=a a^{\prime \prime} \tag{3.1510} \end{align*}Rxtxt=tΓxxtxΓtxt+ΓtαtΓxxαΓxαtΓtxα=tΓxxtΓxxtΓtxx(3.1510)=t(aa)aaaa=(a)2+aa(a)2=aa
Furthermore, we have g t t = 1 , g x x = a 2 g t t = 1 , g x x = a 2 g_(tt)=1,g_(xx)=-a^(2)g_{t t}=1, g_{x x}=-a^{2}gtt=1,gxx=a2, and g t x = g x t = 0 g t x = g x t = 0 g_(tx)=g_(xt)=0g_{t x}=g_{x t}=0gtx=gxt=0, which imply that
(3.1511) g t t = 1 , g x x = 1 a 2 , g t x = g x t = 0 (3.1511) g t t = 1 , g x x = 1 a 2 , g t x = g x t = 0 {:(3.1511)g^(tt)=1","quadg^(xx)=-(1)/(a^(2))","quadg^(tx)=g^(xt)=0:}\begin{equation*} g^{t t}=1, \quad g^{x x}=-\frac{1}{a^{2}}, \quad g^{t x}=g^{x t}=0 \tag{3.1511} \end{equation*}(3.1511)gtt=1,gxx=1a2,gtx=gxt=0
Thus, we find the nonzero components of the Riemann curvature tensor
(3.1512) R t x t x = g t t R x t x t = a a = R x t x t = R t x x t = R x t t x (3.1512) R t x t x = g t t R x t x t = a a = R x t x t = R t x x t = R x t t x {:(3.1512)R_(txtx)=g_(tt)R_(xtx)^(t)=aa^('')=R_(xtxt)=-R_(txxt)=-R_(xttx):}\begin{equation*} R_{t x t x}=g_{t t} R_{x t x}^{t}=a a^{\prime \prime}=R_{x t x t}=-R_{t x x t}=-R_{x t t x} \tag{3.1512} \end{equation*}(3.1512)Rtxtx=gttRxtxt=aa=Rxtxt=Rtxxt=Rxttx
where the last identities follow from general symmetry properties of the Riemann curvature tensor. The other components of the Riemann curvature tensor R μ v α β R μ v α β R_(mu v alpha beta)R_{\mu v \alpha \beta}Rμvαβ are zero due to general symmetry properties. Thus, we obtain the components of the Ricci tensor R μ ν R μ ν R_(mu nu)R_{\mu \nu}Rμν as
(3.1513) R t t = g α β R t α t β = g x x R t x t x = 1 a 2 a a = a a (3.1514) R x x = g α β R x α x β = g t t R x t x t = 1 a a = a a (3.1515) R t x = g α β R t α x β = 0 (3.1516) R x t = g α β R x α t β = 0 (3.1513) R t t = g α β R t α t β = g x x R t x t x = 1 a 2 a a = a a (3.1514) R x x = g α β R x α x β = g t t R x t x t = 1 a a = a a (3.1515) R t x = g α β R t α x β = 0 (3.1516) R x t = g α β R x α t β = 0 {:[(3.1513)R_(tt)=g^(alpha beta)R_(t alpha t beta)=g^(xx)R_(txtx)=-(1)/(a^(2))aa^('')=-(a^(''))/(a)],[(3.1514)R_(xx)=g^(alpha beta)R_(x alpha x beta)=g^(tt)R_(xtxt)=1*aa^('')=aa^('')],[(3.1515)R_(tx)=g^(alpha beta)R_(t alpha x beta)=0],[(3.1516)R_(xt)=g^(alpha beta)R_(x alpha t beta)=0]:}\begin{align*} R_{t t} & =g^{\alpha \beta} R_{t \alpha t \beta}=g^{x x} R_{t x t x}=-\frac{1}{a^{2}} a a^{\prime \prime}=-\frac{a^{\prime \prime}}{a} \tag{3.1513}\\ R_{x x} & =g^{\alpha \beta} R_{x \alpha x \beta}=g^{t t} R_{x t x t}=1 \cdot a a^{\prime \prime}=a a^{\prime \prime} \tag{3.1514}\\ R_{t x} & =g^{\alpha \beta} R_{t \alpha x \beta}=0 \tag{3.1515}\\ R_{x t} & =g^{\alpha \beta} R_{x \alpha t \beta}=0 \tag{3.1516} \end{align*}(3.1513)Rtt=gαβRtαtβ=gxxRtxtx=1a2aa=aa(3.1514)Rxx=gαβRxαxβ=gttRxtxt=1aa=aa(3.1515)Rtx=gαβRtαxβ=0(3.1516)Rxt=gαβRxαtβ=0
which can be summarized as R μ ν = g μ ν a / a R μ ν = g μ ν a / a R_(mu nu)=-g_(mu nu)a^('')//aR_{\mu \nu}=-g_{\mu \nu} a^{\prime \prime} / aRμν=gμνa/a.
Note that the Ricci scalar R g μ ν R μ ν R g μ ν R μ ν R-=g^(mu nu)R_(mu nu)R \equiv g^{\mu \nu} R_{\mu \nu}RgμνRμν is given by
(3.1517) R = g t t R t t + g x x R x x = 1 ( a a ) + ( 1 a 2 ) a a = 2 a a (3.1517) R = g t t R t t + g x x R x x = 1 a a + 1 a 2 a a = 2 a a {:(3.1517)R=g^(tt)R_(tt)+g^(xx)R_(xx)=1*(-(a^(''))/(a))+(-(1)/(a^(2)))aa^('')=-(2a^(''))/(a):}\begin{equation*} R=g^{t t} R_{t t}+g^{x x} R_{x x}=1 \cdot\left(-\frac{a^{\prime \prime}}{a}\right)+\left(-\frac{1}{a^{2}}\right) a a^{\prime \prime}=-\frac{2 a^{\prime \prime}}{a} \tag{3.1517} \end{equation*}(3.1517)R=gttRtt+gxxRxx=1(aa)+(1a2)aa=2aa
and thus, the Einstein tensor G μ ν R μ ν 1 2 R g μ ν G μ ν R μ ν 1 2 R g μ ν G_(mu nu)-=R_(mu nu)-(1)/(2)Rg_(mu nu)G_{\mu \nu} \equiv R_{\mu \nu}-\frac{1}{2} R g_{\mu \nu}GμνRμν12Rgμν vanishes, i.e.,
(3.1518) G μ ν = g μ ν a a 1 2 ( 2 a a ) g μ ν = 0 (3.1518) G μ ν = g μ ν a a 1 2 2 a a g μ ν = 0 {:(3.1518)G_(mu nu)=-g_(mu nu)(a^(''))/(a)-(1)/(2)(-(2a^(''))/(a))g_(mu nu)=0:}\begin{equation*} G_{\mu \nu}=-g_{\mu \nu} \frac{a^{\prime \prime}}{a}-\frac{1}{2}\left(-\frac{2 a^{\prime \prime}}{a}\right) g_{\mu \nu}=0 \tag{3.1518} \end{equation*}(3.1518)Gμν=gμνaa12(2aa)gμν=0
b) We can find the trajectory of a light ray from L = t ˙ 2 a ( t ) 2 x ˙ 2 = 0 L = t ˙ 2 a ( t ) 2 x ˙ 2 = 0 L=t^(˙)^(2)-a(t)^(2)x^(˙)^(2)=0\mathcal{L}=\dot{t}^{2}-a(t)^{2} \dot{x}^{2}=0L=t˙2a(t)2x˙2=0, where t ˙ = d t / d τ , x ˙ = d x / d τ t ˙ = d t / d τ , x ˙ = d x / d τ t^(˙)=dt//d tau,x^(˙)=dx//d tau\dot{t}=d t / d \tau, \dot{x}=d x / d \taut˙=dt/dτ,x˙=dx/dτ, and τ τ tau\tauτ being the proper time, i.e.,
(3.1519) ( d x d t ) 2 = 1 a 2 d x ( t ) d t = 1 a ( t ) = A 2 + B 2 t 2 (3.1519) d x d t 2 = 1 a 2 d x ( t ) d t = 1 a ( t ) = A 2 + B 2 t 2 {:(3.1519)((dx)/(dt))^(2)=(1)/(a^(2))quad=>quad(dx(t))/(dt)=(1)/(a(t))=A^(2)+B^(2)t^(2):}\begin{equation*} \left(\frac{d x}{d t}\right)^{2}=\frac{1}{a^{2}} \quad \Rightarrow \quad \frac{d x(t)}{d t}=\frac{1}{a(t)}=A^{2}+B^{2} t^{2} \tag{3.1519} \end{equation*}(3.1519)(dxdt)2=1a2dx(t)dt=1a(t)=A2+B2t2
where we choose the plus sign of the square root, since it should be assumed that d x / d t > 0 d x / d t > 0 dx//dt > 0d x / d t>0dx/dt>0 at t = 0 t = 0 t=0t=0t=0. Thus, integrating, we obtain the trajectory of the light ray as
(3.1520) x ( t ) = x 0 + A 2 t + B 2 3 t 3 (3.1520) x ( t ) = x 0 + A 2 t + B 2 3 t 3 {:(3.1520)x(t)=x_(0)+A^(2)t+(B^(2))/(3)t^(3):}\begin{equation*} x(t)=x_{0}+A^{2} t+\frac{B^{2}}{3} t^{3} \tag{3.1520} \end{equation*}(3.1520)x(t)=x0+A2t+B23t3
where x 0 = x ( 0 ) x 0 = x ( 0 ) x_(0)=x(0)x_{0}=x(0)x0=x(0).

2.88

The metric tensor for the three-dimensional Robertson-Walker spacetime can be written as
(3.1521) g = ( g μ ν ) = ( g t t g t r g t ϕ g r t g r r g r ϕ g ϕ t g ϕ r g ϕ ϕ ) = ( 1 0 0 0 a 2 1 k r 2 0 0 0 a 2 r 2 ) . (3.1521) g = g μ ν = g t t g t r g t ϕ g r t g r r g r ϕ g ϕ t g ϕ r g ϕ ϕ = 1 0 0 0 a 2 1 k r 2 0 0 0 a 2 r 2 . {:(3.1521)g=(g_(mu nu))=([g_(tt),g_(tr),g_(t phi)],[g_(rt),g_(rr),g_(r phi)],[g_(phi t),g_(phi r),g_(phi phi)])=([1,0,0],[0,-(a^(2))/(1-kr^(2)),0],[0,0,-a^(2)r^(2)]).:}g=\left(g_{\mu \nu}\right)=\left(\begin{array}{lll} g_{t t} & g_{t r} & g_{t \phi} \tag{3.1521}\\ g_{r t} & g_{r r} & g_{r \phi} \\ g_{\phi t} & g_{\phi r} & g_{\phi \phi} \end{array}\right)=\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & -\frac{a^{2}}{1-k r^{2}} & 0 \\ 0 & 0 & -a^{2} r^{2} \end{array}\right) .(3.1521)g=(gμν)=(gttgtrgtϕgrtgrrgrϕgϕtgϕrgϕϕ)=(1000a21kr2000a2r2).
Inserting the Lagrangian
(3.1522) L = g μ ν x ˙ μ x ˙ ν = t ˙ 2 a 2 1 k r 2 r ˙ 2 a 2 r 2 ϕ ˙ 2 (3.1522) L = g μ ν x ˙ μ x ˙ ν = t ˙ 2 a 2 1 k r 2 r ˙ 2 a 2 r 2 ϕ ˙ 2 {:(3.1522)L=g_(mu nu)x^(˙)^(mu)x^(˙)^(nu)=t^(˙)^(2)-(a^(2))/(1-kr^(2))r^(˙)^(2)-a^(2)r^(2)phi^(˙)^(2):}\begin{equation*} \mathcal{L}=g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=\dot{t}^{2}-\frac{a^{2}}{1-k r^{2}} \dot{r}^{2}-a^{2} r^{2} \dot{\phi}^{2} \tag{3.1522} \end{equation*}(3.1522)L=gμνx˙μx˙ν=t˙2a21kr2r˙2a2r2ϕ˙2
into Euler-Lagrange equations, we find that
L t d d τ L t ˙ = 0 , L t = 2 a a 1 k r 2 r ˙ 2 2 a a r 2 ϕ ˙ 2 , L t ˙ = 2 t ˙ , (3.1523) 2 a a 1 k r 2 r ˙ 2 2 a a r 2 ϕ ˙ 2 2 t ¨ = 0 , L r d d τ L r ˙ = 0 , L r = 2 k a 2 r ( 1 k r 2 ) 2 r ˙ 2 2 a 2 r ϕ ˙ 2 , L r ˙ = 2 a 2 1 k r 2 r ˙ , (3.1524) 2 k a 2 r ( 1 k r 2 ) 2 r ˙ 2 2 a 2 r ϕ ˙ 2 + 4 a 1 k r 2 a ˙ r ˙ + 2 a 2 1 k r 2 r ¨ = 0 , L ϕ d d τ L ϕ ˙ = 0 , L ϕ = 0 , L ϕ ˙ = 2 a 2 r 2 ϕ ˙ , (3.1525) 4 a r 2 a ˙ ϕ ˙ + 4 a 2 r r ˙ ϕ ˙ + 2 a 2 r 2 ϕ ¨ = 0 . L t d d τ L t ˙ = 0 , L t = 2 a a 1 k r 2 r ˙ 2 2 a a r 2 ϕ ˙ 2 , L t ˙ = 2 t ˙ , (3.1523) 2 a a 1 k r 2 r ˙ 2 2 a a r 2 ϕ ˙ 2 2 t ¨ = 0 , L r d d τ L r ˙ = 0 , L r = 2 k a 2 r 1 k r 2 2 r ˙ 2 2 a 2 r ϕ ˙ 2 , L r ˙ = 2 a 2 1 k r 2 r ˙ , (3.1524) 2 k a 2 r 1 k r 2 2 r ˙ 2 2 a 2 r ϕ ˙ 2 + 4 a 1 k r 2 a ˙ r ˙ + 2 a 2 1 k r 2 r ¨ = 0 , L ϕ d d τ L ϕ ˙ = 0 , L ϕ = 0 , L ϕ ˙ = 2 a 2 r 2 ϕ ˙ , (3.1525) 4 a r 2 a ˙ ϕ ˙ + 4 a 2 r r ˙ ϕ ˙ + 2 a 2 r 2 ϕ ¨ = 0 . {:[(delL)/(del t)-(d)/(d tau)(delL)/(del(t^(˙)))=0","quad(delL)/(del t)=-(2aa^('))/(1-kr^(2))r^(˙)^(2)-2aa^(')r^(2)phi^(˙)^(2)","quad(delL)/(del(t^(˙)))=2t^(˙)","],[(3.1523)=>quad-(2aa^('))/(1-kr^(2))r^(˙)^(2)-2aa^(')r^(2)phi^(˙)^(2)-2t^(¨)=0","],[(delL)/(del r)-(d)/(d tau)(delL)/(del(r^(˙)))=0","quad(delL)/(del r)=-(2ka^(2)r)/((1-kr^(2))^(2))r^(˙)^(2)-2a^(2)rphi^(˙)^(2)","quad(delL)/(del(r^(˙)))=-(2a^(2))/(1-kr^(2))r^(˙)","],[(3.1524)=>quad(2ka^(2)r)/((1-kr^(2))^(2))r^(˙)^(2)-2a^(2)rphi^(˙)^(2)+(4a)/(1-kr^(2))a^(˙)r^(˙)+(2a^(2))/(1-kr^(2))r^(¨)=0","],[(delL)/(del phi)-(d)/(d tau)(delL)/(del(phi^(˙)))=0","quad(delL)/(del phi)=0","quad(delL)/(del(phi^(˙)))=-2a^(2)r^(2)phi^(˙)","],[(3.1525)=>quad4ar^(2)a^(˙)phi^(˙)+4a^(2)rr^(˙)phi^(˙)+2a^(2)r^(2)phi^(¨)=0.]:}\begin{align*} & \frac{\partial \mathcal{L}}{\partial t}-\frac{d}{d \tau} \frac{\partial \mathcal{L}}{\partial \dot{t}}=0, \quad \frac{\partial \mathcal{L}}{\partial t}=-\frac{2 a a^{\prime}}{1-k r^{2}} \dot{r}^{2}-2 a a^{\prime} r^{2} \dot{\phi}^{2}, \quad \frac{\partial \mathcal{L}}{\partial \dot{t}}=2 \dot{t}, \\ & \Rightarrow \quad-\frac{2 a a^{\prime}}{1-k r^{2}} \dot{r}^{2}-2 a a^{\prime} r^{2} \dot{\phi}^{2}-2 \ddot{t}=0, \tag{3.1523}\\ & \frac{\partial \mathcal{L}}{\partial r}-\frac{d}{d \tau} \frac{\partial \mathcal{L}}{\partial \dot{r}}=0, \quad \frac{\partial \mathcal{L}}{\partial r}=-\frac{2 k a^{2} r}{\left(1-k r^{2}\right)^{2}} \dot{r}^{2}-2 a^{2} r \dot{\phi}^{2}, \quad \frac{\partial \mathcal{L}}{\partial \dot{r}}=-\frac{2 a^{2}}{1-k r^{2}} \dot{r}, \\ & \Rightarrow \quad \frac{2 k a^{2} r}{\left(1-k r^{2}\right)^{2}} \dot{r}^{2}-2 a^{2} r \dot{\phi}^{2}+\frac{4 a}{1-k r^{2}} \dot{a} \dot{r}+\frac{2 a^{2}}{1-k r^{2}} \ddot{r}=0, \tag{3.1524}\\ & \frac{\partial \mathcal{L}}{\partial \phi}-\frac{d}{d \tau} \frac{\partial \mathcal{L}}{\partial \dot{\phi}}=0, \quad \frac{\partial \mathcal{L}}{\partial \phi}=0, \quad \frac{\partial \mathcal{L}}{\partial \dot{\phi}}=-2 a^{2} r^{2} \dot{\phi}, \\ & \Rightarrow \quad 4 a r^{2} \dot{a} \dot{\phi}+4 a^{2} r \dot{r} \dot{\phi}+2 a^{2} r^{2} \ddot{\phi}=0 . \tag{3.1525} \end{align*}LtddτLt˙=0,Lt=2aa1kr2r˙22aar2ϕ˙2,Lt˙=2t˙,(3.1523)2aa1kr2r˙22aar2ϕ˙22t¨=0,LrddτLr˙=0,Lr=2ka2r(1kr2)2r˙22a2rϕ˙2,Lr˙=2a21kr2r˙,(3.1524)2ka2r(1kr2)2r˙22a2rϕ˙2+4a1kr2a˙r˙+2a21kr2r¨=0,LϕddτLϕ˙=0,Lϕ=0,Lϕ˙=2a2r2ϕ˙,(3.1525)4ar2a˙ϕ˙+4a2rr˙ϕ˙+2a2r2ϕ¨=0.
Now, using that
(3.1526) a ˙ = d a d τ = d a d t d t d τ = a t ˙ (3.1526) a ˙ = d a d τ = d a d t d t d τ = a t ˙ {:(3.1526)a^(˙)=(da)/(d tau)=(da)/(dt)(dt)/(d tau)=a^(')t^(˙):}\begin{equation*} \dot{a}=\frac{d a}{d \tau}=\frac{d a}{d t} \frac{d t}{d \tau}=a^{\prime} \dot{t} \tag{3.1526} \end{equation*}(3.1526)a˙=dadτ=dadtdtdτ=at˙
we obtain the geodesic equations as
(3.1527) t ¨ + a a 1 k r 2 r ˙ 2 + a a r 2 ϕ ˙ 2 = 0 (3.1528) r ¨ + 2 a a t ˙ r ˙ + k r 1 k r 2 r ˙ 2 ( 1 k r 2 ) r ϕ ˙ 2 = 0 (3.1529) ϕ ¨ + 2 a a t ˙ ϕ ˙ + 2 r r ˙ ϕ ˙ = 0 (3.1527) t ¨ + a a 1 k r 2 r ˙ 2 + a a r 2 ϕ ˙ 2 = 0 (3.1528) r ¨ + 2 a a t ˙ r ˙ + k r 1 k r 2 r ˙ 2 1 k r 2 r ϕ ˙ 2 = 0 (3.1529) ϕ ¨ + 2 a a t ˙ ϕ ˙ + 2 r r ˙ ϕ ˙ = 0 {:[(3.1527)t^(¨)+(aa^('))/(1-kr^(2))r^(˙)^(2)+aa^(')r^(2)phi^(˙)^(2)=0],[(3.1528)r^(¨)+(2a^('))/(a)t^(˙)r^(˙)+(kr)/(1-kr^(2))r^(˙)^(2)-(1-kr^(2))rphi^(˙)^(2)=0],[(3.1529)phi^(¨)+(2a^('))/(a)t^(˙)phi^(˙)+(2)/(r)r^(˙)phi^(˙)=0]:}\begin{align*} \ddot{t}+\frac{a a^{\prime}}{1-k r^{2}} \dot{r}^{2}+a a^{\prime} r^{2} \dot{\phi}^{2} & =0 \tag{3.1527}\\ \ddot{r}+\frac{2 a^{\prime}}{a} \dot{t} \dot{r}+\frac{k r}{1-k r^{2}} \dot{r}^{2}-\left(1-k r^{2}\right) r \dot{\phi}^{2} & =0 \tag{3.1528}\\ \ddot{\phi}+\frac{2 a^{\prime}}{a} \dot{t} \dot{\phi}+\frac{2}{r} \dot{r} \dot{\phi} & =0 \tag{3.1529} \end{align*}(3.1527)t¨+aa1kr2r˙2+aar2ϕ˙2=0(3.1528)r¨+2aat˙r˙+kr1kr2r˙2(1kr2)rϕ˙2=0(3.1529)ϕ¨+2aat˙ϕ˙+2rr˙ϕ˙=0
Finally, using the general formula for the geodesic equations x ¨ μ + Γ v λ μ x ˙ ν x ˙ λ = 0 x ¨ μ + Γ v λ μ x ˙ ν x ˙ λ = 0 x^(¨)^(mu)+Gamma_(v lambda)^(mu)x^(˙)^(nu)x^(˙)^(lambda)=0\ddot{x}^{\mu}+\Gamma_{v \lambda}^{\mu} \dot{x}^{\nu} \dot{x}^{\lambda}=0x¨μ+Γvλμx˙νx˙λ=0, we identify the nonzero Christoffel symbols as
(3.1530) Γ r r t = a a 1 k r 2 (3.1531) Γ ϕ ϕ t = a a r 2 (3.1530) Γ r r t = a a 1 k r 2 (3.1531) Γ ϕ ϕ t = a a r 2 {:[(3.1530)Gamma_(rr)^(t)=(aa^('))/(1-kr^(2))],[(3.1531)Gamma_(phi phi)^(t)=aa^(')r^(2)]:}\begin{align*} \Gamma_{r r}^{t} & =\frac{a a^{\prime}}{1-k r^{2}} \tag{3.1530}\\ \Gamma_{\phi \phi}^{t} & =a a^{\prime} r^{2} \tag{3.1531} \end{align*}(3.1530)Γrrt=aa1kr2(3.1531)Γϕϕt=aar2
(3.1532) Γ t r r = Γ r t r = a a (3.1533) Γ r r r = k r 1 k r 2 (3.1534) Γ ϕ ϕ r = ( 1 k r 2 ) r (3.1535) Γ t ϕ ϕ = Γ ϕ t ϕ = a a , r (3.1536) Γ r ϕ ϕ = Γ ϕ r ϕ = 1 r (3.1532) Γ t r r = Γ r t r = a a (3.1533) Γ r r r = k r 1 k r 2 (3.1534) Γ ϕ ϕ r = 1 k r 2 r (3.1535) Γ t ϕ ϕ = Γ ϕ t ϕ = a a , r (3.1536) Γ r ϕ ϕ = Γ ϕ r ϕ = 1 r {:[(3.1532)Gamma_(tr)^(r)=Gamma_(rt)^(r)=(a^('))/(a)],[(3.1533)Gamma_(rr)^(r)=(kr)/(1-kr^(2))],[(3.1534)Gamma_(phi phi)^(r)=-(1-kr^(2))r],[(3.1535)Gamma_(t phi)^(phi)=Gamma_(phi t)^(phi)=(a^('))/(a)","r],[(3.1536)Gamma_(r phi)^(phi)=Gamma_(phi r)^(phi)=(1)/(r)]:}\begin{align*} \Gamma_{t r}^{r} & =\Gamma_{r t}^{r}=\frac{a^{\prime}}{a} \tag{3.1532}\\ \Gamma_{r r}^{r} & =\frac{k r}{1-k r^{2}} \tag{3.1533}\\ \Gamma_{\phi \phi}^{r} & =-\left(1-k r^{2}\right) r \tag{3.1534}\\ \Gamma_{t \phi}^{\phi} & =\Gamma_{\phi t}^{\phi}=\frac{a^{\prime}}{a}, r \tag{3.1535}\\ \Gamma_{r \phi}^{\phi} & =\Gamma_{\phi r}^{\phi}=\frac{1}{r} \tag{3.1536} \end{align*}(3.1532)Γtrr=Γrtr=aa(3.1533)Γrrr=kr1kr2(3.1534)Γϕϕr=(1kr2)r(3.1535)Γtϕϕ=Γϕtϕ=aa,r(3.1536)Γrϕϕ=Γϕrϕ=1r

2.89

a) We compute the Euler-Lagrange equation for
(3.1537) L = 1 2 g μ v ( x ( s ) ) x ˙ μ ( s ) x ˙ v ( s ) (3.1537) L = 1 2 g μ v ( x ( s ) ) x ˙ μ ( s ) x ˙ v ( s ) {:(3.1537)L=(1)/(2)g_(mu v)(x(s))x^(˙)^(mu)(s)x^(˙)^(v)(s):}\begin{equation*} \mathcal{L}=\frac{1}{2} g_{\mu v}(x(s)) \dot{x}^{\mu}(s) \dot{x}^{v}(s) \tag{3.1537} \end{equation*}(3.1537)L=12gμv(x(s))x˙μ(s)x˙v(s)
since we know that they are equivalent to
(3.1538) x ¨ μ + Γ α β μ x ˙ α x ˙ β = 0 (3.1538) x ¨ μ + Γ α β μ x ˙ α x ˙ β = 0 {:(3.1538)x^(¨)^(mu)+Gamma_(alpha beta)^(mu)x^(˙)^(alpha)x^(˙)^(beta)=0:}\begin{equation*} \ddot{x}^{\mu}+\Gamma_{\alpha \beta}^{\mu} \dot{x}^{\alpha} \dot{x}^{\beta}=0 \tag{3.1538} \end{equation*}(3.1538)x¨μ+Γαβμx˙αx˙β=0
and thus allow us to read off the nonzero Christoffel symbols in a simple manner. Thus, we obtain
(3.1539) L = 1 2 { t ˙ 2 e 2 t / a [ ( 1 + r 2 / a 2 ) 1 r ˙ 2 + r 2 sin 2 θ φ ˙ 2 + r 2 θ ˙ 2 ] } (3.1539) L = 1 2 t ˙ 2 e 2 t / a 1 + r 2 / a 2 1 r ˙ 2 + r 2 sin 2 θ φ ˙ 2 + r 2 θ ˙ 2 {:(3.1539)L=(1)/(2){t^(˙)^(2)-e^(-2t//a)[(1+r^(2)//a^(2))^(-1)r^(˙)^(2)+r^(2)sin^(2)thetavarphi^(˙)^(2)+r^(2)theta^(˙)^(2)]}:}\begin{equation*} \mathcal{L}=\frac{1}{2}\left\{\dot{t}^{2}-e^{-2 t / a}\left[\left(1+r^{2} / a^{2}\right)^{-1} \dot{r}^{2}+r^{2} \sin ^{2} \theta \dot{\varphi}^{2}+r^{2} \dot{\theta}^{2}\right]\right\} \tag{3.1539} \end{equation*}(3.1539)L=12{t˙2e2t/a[(1+r2/a2)1r˙2+r2sin2θφ˙2+r2θ˙2]}
and the Euler-Lagrange equations
(3.1540) d d s L t ˙ L t = t ¨ 1 a e 2 t / a [ ( 1 + r 2 / a 2 ) 1 r ˙ 2 + r 2 sin 2 θ φ ˙ 2 + r 2 θ ˙ 2 ] = 0 d d s L r ˙ L r = d d s [ e 2 t / a ( 1 + r 2 / a 2 ) 1 r ˙ ] + e 2 t / a [ ( 1 + r 2 / a 2 ) 2 r a 2 r ˙ 2 + r sin 2 θ φ ˙ 2 + r θ ˙ 2 ] = e 2 t / a 1 + r 2 / a 2 [ r ¨ 2 a t ˙ r ˙ r a 2 + r 2 r ˙ 2 (3.1541) d d s L θ ˙ L θ = d d s ( e 2 t / a r 2 θ ˙ ) + e 2 / a 2 ) ( r sin 2 θ φ ˙ 2 + r θ ˙ 2 ) ] = 0 = r 2 ( θ ¨ 2 a t ˙ θ ˙ + 2 r r ˙ θ ˙ sin θ sin θ cos θ φ ˙ 2 (3.1540) d d s L t ˙ L t = t ¨ 1 a e 2 t / a 1 + r 2 / a 2 1 r ˙ 2 + r 2 sin 2 θ φ ˙ 2 + r 2 θ ˙ 2 = 0 d d s L r ˙ L r = d d s e 2 t / a 1 + r 2 / a 2 1 r ˙ + e 2 t / a 1 + r 2 / a 2 2 r a 2 r ˙ 2 + r sin 2 θ φ ˙ 2 + r θ ˙ 2 = e 2 t / a 1 + r 2 / a 2 r ¨ 2 a t ˙ r ˙ r a 2 + r 2 r ˙ 2 (3.1541) d d s L θ ˙ L θ = d d s e 2 t / a r 2 θ ˙ + e 2 / a 2 r sin 2 θ φ ˙ 2 + r θ ˙ 2 = 0 = r 2 θ ¨ 2 a t ˙ θ ˙ + 2 r r ˙ θ ˙ sin θ sin θ cos θ φ ˙ 2 {:[(3.1540)(d)/(ds)(delL)/(del(t^(˙)))-(delL)/(del t)=t^(¨)-(1)/(a)e^(-2t//a)[(1+r^(2)//a^(2))^(-1)r^(˙)^(2)+r^(2)sin^(2)thetavarphi^(˙)^(2)+r^(2)theta^(˙)^(2)]=0],[(d)/(ds)(delL)/(del(r^(˙)))-(delL)/(del r)=-(d)/(ds)[e^(-2t//a)(1+r^(2)//a^(2))^(-1)(r^(˙))]],[+e^(-2t//a)[-(1+r^(2)//a^(2))^(-2)(r)/(a^(2))r^(˙)^(2)+rsin^(2)thetavarphi^(˙)^(2)+rtheta^(˙)^(2)]],[=-(e^(-2t//a))/(1+r^(2)//a^(2))[(r^(¨))-(2)/(a)(t^(˙))(r^(˙))-(r)/(a^(2)+r^(2))r^(˙)^(2):}],[(3.1541)(d)/(ds)(delL)/(del(theta^(˙)))-(delL)/(del theta)={:-(d)/(ds)(e^(-2t//a)r^(2)(theta^(˙)))+e^(2)//a^(2))(rsin^(2)thetavarphi^(˙)^(2)+rtheta^(˙)^(2))]=0],[=-r^(2)((theta^(¨))-(2)/(a)(t^(˙))(theta^(˙))+(2)/(r)(r^(˙))(theta^(˙))-sin theta sin theta cos thetavarphi^(˙)^(2):}]:}\begin{align*} \frac{d}{d s} \frac{\partial \mathcal{L}}{\partial \dot{t}}-\frac{\partial \mathcal{L}}{\partial t}= & \ddot{t}-\frac{1}{a} e^{-2 t / a}\left[\left(1+r^{2} / a^{2}\right)^{-1} \dot{r}^{2}+r^{2} \sin ^{2} \theta \dot{\varphi}^{2}+r^{2} \dot{\theta}^{2}\right]=0 \tag{3.1540}\\ \frac{d}{d s} \frac{\partial \mathcal{L}}{\partial \dot{r}}-\frac{\partial \mathcal{L}}{\partial r}= & -\frac{d}{d s}\left[e^{-2 t / a}\left(1+r^{2} / a^{2}\right)^{-1} \dot{r}\right] \\ & +e^{-2 t / a}\left[-\left(1+r^{2} / a^{2}\right)^{-2} \frac{r}{a^{2}} \dot{r}^{2}+r \sin ^{2} \theta \dot{\varphi}^{2}+r \dot{\theta}^{2}\right] \\ = & -\frac{e^{-2 t / a}}{1+r^{2} / a^{2}}\left[\ddot{r}-\frac{2}{a} \dot{t} \dot{r}-\frac{r}{a^{2}+r^{2}} \dot{r}^{2}\right. \\ \frac{d}{d s} \frac{\partial \mathcal{L}}{\partial \dot{\theta}}-\frac{\partial \mathcal{L}}{\partial \theta}= & \left.\left.-\frac{d}{d s}\left(e^{-2 t / a} r^{2} \dot{\theta}\right)+e^{2} / a^{2}\right)\left(r \sin ^{2} \theta \dot{\varphi}^{2}+r \dot{\theta}^{2}\right)\right]=0 \tag{3.1541}\\ = & -r^{2}\left(\ddot{\theta}-\frac{2}{a} \dot{t} \dot{\theta}+\frac{2}{r} \dot{r} \dot{\theta}-\sin \theta \sin \theta \cos \theta \dot{\varphi}^{2}\right. \end{align*}(3.1540)ddsLt˙Lt=t¨1ae2t/a[(1+r2/a2)1r˙2+r2sin2θφ˙2+r2θ˙2]=0ddsLr˙Lr=dds[e2t/a(1+r2/a2)1r˙]+e2t/a[(1+r2/a2)2ra2r˙2+rsin2θφ˙2+rθ˙2]=e2t/a1+r2/a2[r¨2at˙r˙ra2+r2r˙2(3.1541)ddsLθ˙Lθ=dds(e2t/ar2θ˙)+e2/a2)(rsin2θφ˙2+rθ˙2)]=0=r2(θ¨2at˙θ˙+2rr˙θ˙sinθsinθcosθφ˙2
d d s L φ ˙ L φ = d d s ( e 2 t / a r 2 sin 2 θ φ ˙ ) (3.1543) = e 2 t / a r 2 sin 2 θ ( φ ¨ 2 a φ ˙ t ˙ + 2 r r ˙ φ ˙ + 2 cot θ θ ˙ φ ˙ ) = 0 d d s L φ ˙ L φ = d d s e 2 t / a r 2 sin 2 θ φ ˙ (3.1543) = e 2 t / a r 2 sin 2 θ φ ¨ 2 a φ ˙ t ˙ + 2 r r ˙ φ ˙ + 2 cot θ θ ˙ φ ˙ = 0 {:[(d)/(ds)(delL)/(del(varphi^(˙)))-(delL)/(del varphi)=-(d)/(ds)(e^(-2t//a)r^(2)sin^(2)theta(varphi^(˙)))],[(3.1543)=-e^(-2t//a)r^(2)sin^(2)theta((varphi^(¨))-(2)/(a)(varphi^(˙))(t^(˙))+(2)/(r)(r^(˙))(varphi^(˙))+2cot theta(theta^(˙))(varphi^(˙)))=0]:}\begin{align*} \frac{d}{d s} \frac{\partial \mathcal{L}}{\partial \dot{\varphi}}-\frac{\partial \mathcal{L}}{\partial \varphi} & =-\frac{d}{d s}\left(e^{-2 t / a} r^{2} \sin ^{2} \theta \dot{\varphi}\right) \\ & =-e^{-2 t / a} r^{2} \sin ^{2} \theta\left(\ddot{\varphi}-\frac{2}{a} \dot{\varphi} \dot{t}+\frac{2}{r} \dot{r} \dot{\varphi}+2 \cot \theta \dot{\theta} \dot{\varphi}\right)=0 \tag{3.1543} \end{align*}ddsLφ˙Lφ=dds(e2t/ar2sin2θφ˙)(3.1543)=e2t/ar2sin2θ(φ¨2aφ˙t˙+2rr˙φ˙+2cotθθ˙φ˙)=0
from which we find the nonzero Christoffel symbols Γ α β μ = Γ β α μ Γ α β μ = Γ β α μ Gamma_(alpha beta)^(mu)=Gamma_(beta alpha)^(mu)\Gamma_{\alpha \beta}^{\mu}=\Gamma_{\beta \alpha}^{\mu}Γαβμ=Γβαμ such that
Γ r r t = 1 a e 2 t / a ( 1 + r 2 / a 2 ) 1 , Γ θ θ t = 1 a e 2 t / a r 2 , Γ φ φ t = 1 a e 2 t / a r 2 sin 2 θ Γ r r t = 1 a e 2 t / a 1 + r 2 / a 2 1 , Γ θ θ t = 1 a e 2 t / a r 2 , Γ φ φ t = 1 a e 2 t / a r 2 sin 2 θ Gamma_(rr)^(t)=-(1)/(a)e^(-2t//a)(1+r^(2)//a^(2))^(-1),quadGamma_(theta theta)^(t)=-(1)/(a)e^(-2t//a)r^(2),quadGamma_(varphi varphi)^(t)=-(1)/(a)e^(-2t//a)r^(2)sin^(2)theta\Gamma_{r r}^{t}=-\frac{1}{a} e^{-2 t / a}\left(1+r^{2} / a^{2}\right)^{-1}, \quad \Gamma_{\theta \theta}^{t}=-\frac{1}{a} e^{-2 t / a} r^{2}, \quad \Gamma_{\varphi \varphi}^{t}=-\frac{1}{a} e^{-2 t / a} r^{2} \sin ^{2} \thetaΓrrt=1ae2t/a(1+r2/a2)1,Γθθt=1ae2t/ar2,Γφφt=1ae2t/ar2sin2θ,
Γ r t r = 1 a , Γ r r r = r a 2 + r 2 Γ r t r = 1 a , Γ r r r = r a 2 + r 2 Gamma_(rt)^(r)=-(1)/(a),quadGamma_(rr)^(r)=-(r)/(a^(2)+r^(2))\Gamma_{r t}^{r}=-\frac{1}{a}, \quad \Gamma_{r r}^{r}=-\frac{r}{a^{2}+r^{2}}Γrtr=1a,Γrrr=ra2+r2,
Γ θ θ r = ( 1 + r 2 / a 2 ) r , Γ φ φ r = ( 1 + r 2 / a 2 ) r sin 2 θ Γ θ θ r = 1 + r 2 / a 2 r , Γ φ φ r = 1 + r 2 / a 2 r sin 2 θ Gamma_(theta theta)^(r)=-(1+r^(2)//a^(2))r,quadGamma_(varphi varphi)^(r)=-(1+r^(2)//a^(2))rsin^(2)theta\Gamma_{\theta \theta}^{r}=-\left(1+r^{2} / a^{2}\right) r, \quad \Gamma_{\varphi \varphi}^{r}=-\left(1+r^{2} / a^{2}\right) r \sin ^{2} \thetaΓθθr=(1+r2/a2)r,Γφφr=(1+r2/a2)rsin2θ,
Γ t θ θ = 1 a , Γ r θ θ = 1 r , Γ φ φ θ = sin θ cos θ Γ t θ θ = 1 a , Γ r θ θ = 1 r , Γ φ φ θ = sin θ cos θ Gamma_(t theta)^(theta)=-(1)/(a),quadGamma_(r theta)^(theta)=(1)/(r),quadGamma_(varphi varphi)^(theta)=-sin theta cos theta\Gamma_{t \theta}^{\theta}=-\frac{1}{a}, \quad \Gamma_{r \theta}^{\theta}=\frac{1}{r}, \quad \Gamma_{\varphi \varphi}^{\theta}=-\sin \theta \cos \thetaΓtθθ=1a,Γrθθ=1r,Γφφθ=sinθcosθ,
Γ t φ φ = 1 a , Γ r φ φ = 1 r , Γ θ φ φ = cot θ Γ t φ φ = 1 a , Γ r φ φ = 1 r , Γ θ φ φ = cot θ Gamma_(t varphi)^(varphi)=-(1)/(a),quadGamma_(r varphi)^(varphi)=(1)/(r),quadGamma_(theta varphi)^(varphi)=cot theta\Gamma_{t \varphi}^{\varphi}=-\frac{1}{a}, \quad \Gamma_{r \varphi}^{\varphi}=\frac{1}{r}, \quad \Gamma_{\theta \varphi}^{\varphi}=\cot \thetaΓtφφ=1a,Γrφφ=1r,Γθφφ=cotθ.
We now compute the covariant derivative μ A v μ A v grad_(mu)A^(v)\nabla_{\mu} A^{v}μAv of the vector field A v A v A^(v)A^{v}Av
(3.1545) μ A v = μ A v + Γ μ α v A α (3.1545) μ A v = μ A v + Γ μ α v A α {:(3.1545)grad_(mu)A^(v)=del_(mu)A^(v)+Gamma_(mu alpha)^(v)A^(alpha):}\begin{equation*} \nabla_{\mu} A^{v}=\partial_{\mu} A^{v}+\Gamma_{\mu \alpha}^{v} A^{\alpha} \tag{3.1545} \end{equation*}(3.1545)μAv=μAv+ΓμαvAα
i.e., in component form, we obtain
(3.1546) t A t = t A t + Γ t t t A t + Γ t r t A r = 1 a (3.1547) r A t = r A t + Γ r t t A t + Γ r r t A r = r a 2 e 2 t / a ( 1 + r 2 / a 2 ) 1 (3.1548) t A r = t A r + Γ t t r A t + Γ t r r A r = r a 2 (3.1549) r A r = r A r + Γ r t r A t + Γ r r r A r = 1 a t a 2 1 a r 2 / a 2 1 + r 2 / a 2 (3.1546) t A t = t A t + Γ t t t A t + Γ t r t A r = 1 a (3.1547) r A t = r A t + Γ r t t A t + Γ r r t A r = r a 2 e 2 t / a 1 + r 2 / a 2 1 (3.1548) t A r = t A r + Γ t t r A t + Γ t r r A r = r a 2 (3.1549) r A r = r A r + Γ r t r A t + Γ r r r A r = 1 a t a 2 1 a r 2 / a 2 1 + r 2 / a 2 {:[(3.1546)grad_(t)A^(t)=del_(t)A^(t)+Gamma_(tt)^(t)A^(t)+Gamma_(tr)^(t)A^(r)=(1)/(a)],[(3.1547)grad_(r)A^(t)=del_(r)A^(t)+Gamma_(rt)^(t)A^(t)+Gamma_(rr)^(t)A^(r)=-(r)/(a^(2))e^(-2t//a)(1+r^(2)//a^(2))^(-1)],[(3.1548)grad_(t)A^(r)=del_(t)A^(r)+Gamma_(tt)^(r)A^(t)+Gamma_(tr)^(r)A^(r)=-(r)/(a^(2))],[(3.1549)grad_(r)A^(r)=del_(r)A^(r)+Gamma_(rt)^(r)A^(t)+Gamma_(rr)^(r)A^(r)=(1)/(a)-(t)/(a^(2))-(1)/(a)(r^(2)//a^(2))/(1+r^(2)//a^(2))]:}\begin{align*} & \nabla_{t} A^{t}=\partial_{t} A^{t}+\Gamma_{t t}^{t} A^{t}+\Gamma_{t r}^{t} A^{r}=\frac{1}{a} \tag{3.1546}\\ & \nabla_{r} A^{t}=\partial_{r} A^{t}+\Gamma_{r t}^{t} A^{t}+\Gamma_{r r}^{t} A^{r}=-\frac{r}{a^{2}} e^{-2 t / a}\left(1+r^{2} / a^{2}\right)^{-1} \tag{3.1547}\\ & \nabla_{t} A^{r}=\partial_{t} A^{r}+\Gamma_{t t}^{r} A^{t}+\Gamma_{t r}^{r} A^{r}=-\frac{r}{a^{2}} \tag{3.1548}\\ & \nabla_{r} A^{r}=\partial_{r} A^{r}+\Gamma_{r t}^{r} A^{t}+\Gamma_{r r}^{r} A^{r}=\frac{1}{a}-\frac{t}{a^{2}}-\frac{1}{a} \frac{r^{2} / a^{2}}{1+r^{2} / a^{2}} \tag{3.1549} \end{align*}(3.1546)tAt=tAt+ΓtttAt+ΓtrtAr=1a(3.1547)rAt=rAt+ΓrttAt+ΓrrtAr=ra2e2t/a(1+r2/a2)1(3.1548)tAr=tAr+ΓttrAt+ΓtrrAr=ra2(3.1549)rAr=rAr+ΓrtrAt+ΓrrrAr=1ata21ar2/a21+r2/a2
b) We have the conservation law
(3.1550) g μ ν ( x ( s ) ) x ˙ μ ( s ) x ˙ ν ( s ) = t ˙ 2 e 2 t / a ( 1 + r 2 / a 2 ) 1 r ˙ 2 = 0 (3.1550) g μ ν ( x ( s ) ) x ˙ μ ( s ) x ˙ ν ( s ) = t ˙ 2 e 2 t / a 1 + r 2 / a 2 1 r ˙ 2 = 0 {:(3.1550)g_(mu nu)(x(s))x^(˙)^(mu)(s)x^(˙)^(nu)(s)=t^(˙)^(2)-e^(-2t//a)(1+r^(2)//a^(2))^(-1)r^(˙)^(2)=0:}\begin{equation*} g_{\mu \nu}(x(s)) \dot{x}^{\mu}(s) \dot{x}^{\nu}(s)=\dot{t}^{2}-e^{-2 t / a}\left(1+r^{2} / a^{2}\right)^{-1} \dot{r}^{2}=0 \tag{3.1550} \end{equation*}(3.1550)gμν(x(s))x˙μ(s)x˙ν(s)=t˙2e2t/a(1+r2/a2)1r˙2=0
since the tangent should be a lightlike vector. From this, we conclude that
(3.1551) d r d t = r ˙ t ¨ = 1 + r 2 / a 2 e t / a (3.1551) d r d t = r ˙ t ¨ = 1 + r 2 / a 2 e t / a {:(3.1551)(dr)/(dt)=((r^(˙)))/((t^(¨)))=sqrt(1+r^(2)//a^(2))e^(t//a):}\begin{equation*} \frac{d r}{d t}=\frac{\dot{r}}{\ddot{t}}=\sqrt{1+r^{2} / a^{2}} e^{t / a} \tag{3.1551} \end{equation*}(3.1551)drdt=r˙t¨=1+r2/a2et/a
with r ( 0 ) = a r ( 0 ) = a r(0)=ar(0)=ar(0)=a. This can be solved by separation
(3.1552) a r ( t ) d r 1 + r 2 / a 2 = 0 t e t / a d t (3.1552) a r ( t ) d r 1 + r 2 / a 2 = 0 t e t / a d t {:(3.1552)int_(a)^(r(t))(dr)/(sqrt(1+r^(2)//a^(2)))=int_(0)^(t)e^(t//a)dt:}\begin{equation*} \int_{a}^{r(t)} \frac{d r}{\sqrt{1+r^{2} / a^{2}}}=\int_{0}^{t} e^{t / a} d t \tag{3.1552} \end{equation*}(3.1552)ar(t)dr1+r2/a2=0tet/adt
i.e.,
(3.1553) a [ arsinh ( r / a ) arsinh ( 1 ) ] = a ( e t / a 1 ) (3.1553) a [ arsinh ( r / a ) arsinh ( 1 ) ] = a e t / a 1 {:(3.1553)a[arsinh(r//a)-arsinh(1)]=a(e^(t//a)-1):}\begin{equation*} a[\operatorname{arsinh}(r / a)-\operatorname{arsinh}(1)]=a\left(e^{t / a}-1\right) \tag{3.1553} \end{equation*}(3.1553)a[arsinh(r/a)arsinh(1)]=a(et/a1)
We thus can obtain the following explicit trajectory of the light pulse
(3.1554) r ( t ) = a sinh ( e t / a + arsinh ( 1 ) 1 ) (3.1554) r ( t ) = a sinh e t / a + arsinh ( 1 ) 1 {:(3.1554)r(t)=a sinh(e^(t//a)+arsinh(1)-1):}\begin{equation*} r(t)=a \sinh \left(e^{t / a}+\operatorname{arsinh}(1)-1\right) \tag{3.1554} \end{equation*}(3.1554)r(t)=asinh(et/a+arsinh(1)1)
2.90
For k = 1 k = 1 k=1k=1k=1 and d Ω = 0 d Ω = 0 d Omega=0d \Omega=0dΩ=0, we can write the Robertson-Walker metric as
(3.1555) d s 2 = c 2 d t 2 S ( t ) 2 d χ 2 = [ c 2 t ˙ 2 S ( t ) 2 χ ˙ 2 ] d τ 2 (3.1555) d s 2 = c 2 d t 2 S ( t ) 2 d χ 2 = c 2 t ˙ 2 S ( t ) 2 χ ˙ 2 d τ 2 {:(3.1555)ds^(2)=c^(2)dt^(2)-S(t)^(2)dchi^(2)=[c^(2)t^(˙)^(2)-S(t)^(2)chi^(˙)^(2)]dtau^(2):}\begin{equation*} d s^{2}=c^{2} d t^{2}-S(t)^{2} d \chi^{2}=\left[c^{2} \dot{t}^{2}-S(t)^{2} \dot{\chi}^{2}\right] d \tau^{2} \tag{3.1555} \end{equation*}(3.1555)ds2=c2dt2S(t)2dχ2=[c2t˙2S(t)2χ˙2]dτ2
where χ = arcsin r χ = arcsin r chi=arcsin r\chi=\arcsin rχ=arcsinr and dot is differentiation with respect to the parameter τ τ tau\tauτ. Thus, we have the metric condition c 2 t ˙ 2 S ( t ) 2 χ ˙ 2 = β = c 2 t ˙ 2 S ( t ) 2 χ ˙ 2 = β = c^(2)t^(˙)^(2)-S(t)^(2)chi^(˙)^(2)=beta=c^{2} \dot{t}^{2}-S(t)^{2} \dot{\chi}^{2}=\beta=c2t˙2S(t)2χ˙2=β= const. for an affinely parametrized geodesic.
a) The Euler-Lagrange equations yield the geodesic equations for t t ttt and χ χ chi\chiχ, i.e.,
(3.1556) t ¨ + 1 c 2 S ( t ) S ( t ) χ ˙ 2 = 0 (3.1557) d d τ [ S ( t ) 2 χ ˙ ] = 0 (3.1556) t ¨ + 1 c 2 S ( t ) S ( t ) χ ˙ 2 = 0 (3.1557) d d τ S ( t ) 2 χ ˙ = 0 {:[(3.1556)t^(¨)+(1)/(c^(2))S(t)S^(')(t)chi^(˙)^(2)=0],[(3.1557)(d)/(d tau)[S(t)^(2)(chi^(˙))]=0]:}\begin{align*} \ddot{t}+\frac{1}{c^{2}} S(t) S^{\prime}(t) \dot{\chi}^{2} & =0 \tag{3.1556}\\ \frac{d}{d \tau}\left[S(t)^{2} \dot{\chi}\right] & =0 \tag{3.1557} \end{align*}(3.1556)t¨+1c2S(t)S(t)χ˙2=0(3.1557)ddτ[S(t)2χ˙]=0
See also the solution to Problem 2.81. Therefore, the first integral is
(3.1558) S ( t ) 2 χ ˙ = α = const. χ ˙ = α S ( t ) 2 (3.1558) S ( t ) 2 χ ˙ = α =  const.  χ ˙ = α S ( t ) 2 {:(3.1558)S(t)^(2)chi^(˙)=alpha=" const. "quad=>quadchi^(˙)=(alpha)/(S(t)^(2)):}\begin{equation*} S(t)^{2} \dot{\chi}=\alpha=\text { const. } \quad \Rightarrow \quad \dot{\chi}=\frac{\alpha}{S(t)^{2}} \tag{3.1558} \end{equation*}(3.1558)S(t)2χ˙=α= const. χ˙=αS(t)2
which from the metric condition gives
(3.1559) t ˙ = ± 1 c β + α 2 S ( t ) 2 (3.1559) t ˙ = ± 1 c β + α 2 S ( t ) 2 {:(3.1559)t^(˙)=+-(1)/(c)sqrt(beta+(alpha^(2))/(S(t)^(2))):}\begin{equation*} \dot{t}= \pm \frac{1}{c} \sqrt{\beta+\frac{\alpha^{2}}{S(t)^{2}}} \tag{3.1559} \end{equation*}(3.1559)t˙=±1cβ+α2S(t)2
b) For lightlike geodesics, β = 0 β = 0 beta=0\beta=0β=0. Thus, inserting β = 0 β = 0 beta=0\beta=0β=0 and α = S ( t ) 2 χ ˙ α = S ( t ) 2 χ ˙ alpha=S(t)^(2)chi^(˙)\alpha=S(t)^{2} \dot{\chi}α=S(t)2χ˙ into the result of a), we find that
(3.1560) i ˙ = ± 1 c α S ( t ) = ± 1 c S ( t ) χ ˙ χ ˙ t ˙ = ± c S ( t ) d χ d t = ± c S ( t ) (3.1560) i ˙ = ± 1 c α S ( t ) = ± 1 c S ( t ) χ ˙ χ ˙ t ˙ = ± c S ( t ) d χ d t = ± c S ( t ) {:(3.1560)i^(˙)=+-(1)/(c)(alpha)/(S(t))=+-(1)/(c)S(t)chi^(˙)quad=>quad((chi^(˙)))/((t^(˙)))=+-(c)/(S(t))quad=>quad(d chi)/(dt)=+-(c)/(S(t)):}\begin{equation*} \dot{i}= \pm \frac{1}{c} \frac{\alpha}{S(t)}= \pm \frac{1}{c} S(t) \dot{\chi} \quad \Rightarrow \quad \frac{\dot{\chi}}{\dot{t}}= \pm \frac{c}{S(t)} \quad \Rightarrow \quad \frac{d \chi}{d t}= \pm \frac{c}{S(t)} \tag{3.1560} \end{equation*}(3.1560)i˙=±1cαS(t)=±1cS(t)χ˙χ˙t˙=±cS(t)dχdt=±cS(t)
Now, using the subsidiary condition on χ χ chi\chiχ given in the problem and assuming propagation forward in time, we have the positive derivative, i.e.,
(3.1561) d χ d t = c S ( t ) d χ = c S ( t ) d t (3.1561) d χ d t = c S ( t ) d χ = c S ( t ) d t {:(3.1561)(d chi)/(dt)=(c)/(S(t))quad=>quad d chi=(c)/(S(t))dt:}\begin{equation*} \frac{d \chi}{d t}=\frac{c}{S(t)} \quad \Rightarrow \quad d \chi=\frac{c}{S(t)} d t \tag{3.1561} \end{equation*}(3.1561)dχdt=cS(t)dχ=cS(t)dt
Integrating leads to
(3.1562) Δ χ χ 1 χ 0 = d χ = t 0 t 1 c S ( t ) d t = 0 T c S ( t + t 0 ) d t (3.1562) Δ χ χ 1 χ 0 = d χ = t 0 t 1 c S ( t ) d t = 0 T c S t + t 0 d t {:(3.1562)Delta chi-=chi_(1)-chi_(0)=int d chi=int_(t_(0))^(t_(1))(c)/(S(t))dt=int_(0)^(T)(c)/(S(t+t_(0)))dt:}\begin{equation*} \Delta \chi \equiv \chi_{1}-\chi_{0}=\int d \chi=\int_{t_{0}}^{t_{1}} \frac{c}{S(t)} d t=\int_{0}^{T} \frac{c}{S\left(t+t_{0}\right)} d t \tag{3.1562} \end{equation*}(3.1562)Δχχ1χ0=dχ=t0t1cS(t)dt=0TcS(t+t0)dt
where t 1 t 0 + T t 1 t 0 + T t_(1)-=t_(0)+Tt_{1} \equiv t_{0}+Tt1t0+T. The coordinate transformation between χ χ chi\chiχ and r r rrr is given by χ = arcsin r χ = arcsin r chi=arcsin r\chi=\arcsin rχ=arcsinr, which means that for r = r 0 = 0 r = r 0 = 0 r=r_(0)=0r=r_{0}=0r=r0=0, we have χ = χ 0 = 0 χ = χ 0 = 0 chi=chi_(0)=0\chi=\chi_{0}=0χ=χ0=0, and thus, we find that
Δ χ ( T ) = 0 T c S ( t + t 0 ) d t arcsin Δ r ( T ) (3.1563) Δ r ( T ) = sin Δ χ ( T ) = sin ( 0 T c S ( t + t 0 ) d t ) Δ χ ( T ) = 0 T c S t + t 0 d t arcsin Δ r ( T ) (3.1563) Δ r ( T ) = sin Δ χ ( T ) = sin 0 T c S t + t 0 d t {:[Delta chi(T)=int_(0)^(T)(c)/(S(t+t_(0)))dt-=arcsin Delta r(T)],[(3.1563)quad<=>quad Delta r(T)=sin Delta chi(T)=sin(int_(0)^(T)(c)/(S(t+t_(0)))dt)]:}\begin{align*} & \Delta \chi(T)=\int_{0}^{T} \frac{c}{S\left(t+t_{0}\right)} d t \equiv \arcsin \Delta r(T) \\ & \quad \Leftrightarrow \quad \Delta r(T)=\sin \Delta \chi(T)=\sin \left(\int_{0}^{T} \frac{c}{S\left(t+t_{0}\right)} d t\right) \tag{3.1563} \end{align*}Δχ(T)=0TcS(t+t0)dtarcsinΔr(T)(3.1563)Δr(T)=sinΔχ(T)=sin(0TcS(t+t0)dt)
Therefore, the distance in the r r rrr-coordinate of the Robertson-Walker metric that a light ray (emitted at universal time t = t 0 t = t 0 t=t_(0)t=t_{0}t=t0 at r = r 0 = 0 r = r 0 = 0 r=r_(0)=0r=r_{0}=0r=r0=0 ) travels in the universal time interval [ t 0 , t 0 + T ] t 0 , t 0 + T [t_(0),t_(0)+T]\left[t_{0}, t_{0}+T\right][t0,t0+T] is given by
(3.1564) Δ r ( T ) = sin ( 0 T c S ( t + t 0 ) d t ) (3.1564) Δ r ( T ) = sin 0 T c S t + t 0 d t {:(3.1564)Delta r(T)=sin(int_(0)^(T)(c)/(S(t+t_(0)))dt):}\begin{equation*} \Delta r(T)=\sin \left(\int_{0}^{T} \frac{c}{S\left(t+t_{0}\right)} d t\right) \tag{3.1564} \end{equation*}(3.1564)Δr(T)=sin(0TcS(t+t0)dt)

2.91

The given standard Schwarzschild metric is
(3.1565) d s 2 = ( 1 r r ) c 2 d t 2 ( 1 r r ) 1 d r 2 r 2 d Ω 2 (3.1565) d s 2 = 1 r r c 2 d t 2 1 r r 1 d r 2 r 2 d Ω 2 {:(3.1565)ds^(2)=(1-(r_(**))/(r))c^(2)dt^(2)-(1-(r_(**))/(r))^(-1)dr^(2)-r^(2)dOmega^(2):}\begin{equation*} d s^{2}=\left(1-\frac{r_{*}}{r}\right) c^{2} d t^{2}-\left(1-\frac{r_{*}}{r}\right)^{-1} d r^{2}-r^{2} d \Omega^{2} \tag{3.1565} \end{equation*}(3.1565)ds2=(1rr)c2dt2(1rr)1dr2r2dΩ2
where s s sss is the path parameter, r 2 G M / c 2 r 2 G M / c 2 r_(**)-=2GM//c^(2)r_{*} \equiv 2 G M / c^{2}r2GM/c2, and d Ω 2 = d θ 2 + sin 2 θ d ϕ 2 d Ω 2 = d θ 2 + sin 2 θ d ϕ 2 dOmega^(2)=dtheta^(2)+sin^(2)theta dphi^(2)d \Omega^{2}=d \theta^{2}+\sin ^{2} \theta d \phi^{2}dΩ2=dθ2+sin2θdϕ2. In addition, the given initial conditions are
(3.1566) r = r 0 , t ˙ = β , r ˙ = α , θ ˙ = 0 , ϕ ˙ = 0 (3.1566) r = r 0 , t ˙ = β , r ˙ = α , θ ˙ = 0 , ϕ ˙ = 0 {:(3.1566)r=r_(0)","quadt^(˙)=beta","quadr^(˙)=alpha","quadtheta^(˙)=0","quadphi^(˙)=0:}\begin{equation*} r=r_{0}, \quad \dot{t}=\beta, \quad \dot{r}=\alpha, \quad \dot{\theta}=0, \quad \dot{\phi}=0 \tag{3.1566} \end{equation*}(3.1566)r=r0,t˙=β,r˙=α,θ˙=0,ϕ˙=0
We introduce the Lagrangian
(3.1567) L = g μ ν x ˙ μ x ˙ ν = ( 1 r r ) c 2 t ˙ 2 ( 1 r r ) 1 r ˙ 2 r 2 θ ˙ 2 r 2 sin 2 θ ϕ ˙ 2 (3.1567) L = g μ ν x ˙ μ x ˙ ν = 1 r r c 2 t ˙ 2 1 r r 1 r ˙ 2 r 2 θ ˙ 2 r 2 sin 2 θ ϕ ˙ 2 {:(3.1567)L=g_(mu nu)x^(˙)^(mu)x^(˙)^(nu)=(1-(r_(**))/(r))c^(2)t^(˙)^(2)-(1-(r_(**))/(r))^(-1)r^(˙)^(2)-r^(2)theta^(˙)^(2)-r^(2)sin^(2)thetaphi^(˙)^(2):}\begin{equation*} \mathcal{L}=g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=\left(1-\frac{r_{*}}{r}\right) c^{2} \dot{t}^{2}-\left(1-\frac{r_{*}}{r}\right)^{-1} \dot{r}^{2}-r^{2} \dot{\theta}^{2}-r^{2} \sin ^{2} \theta \dot{\phi}^{2} \tag{3.1567} \end{equation*}(3.1567)L=gμνx˙μx˙ν=(1rr)c2t˙2(1rr)1r˙2r2θ˙2r2sin2θϕ˙2
Since L t = 0 L t = 0 (delL)/(del t)=0\frac{\partial \mathcal{L}}{\partial t}=0Lt=0 and L ϕ = 0 L ϕ = 0 (delL)/(del phi)=0\frac{\partial \mathcal{L}}{\partial \phi}=0Lϕ=0, we have two constants of motion stemming from the Euler-Lagrange equations for t t ttt and ϕ ϕ phi\phiϕ, namely
(3.1568) d d s [ 2 ( 1 r r ) c 2 t ˙ ] = 0 2 ( 1 r r ) c 2 t ˙ = E = const., , 2 r 2 sin 2 θ ϕ ˙ = L = const. (3.1568) d d s 2 1 r r c 2 t ˙ = 0 2 1 r r c 2 t ˙ = E =  const.,  , 2 r 2 sin 2 θ ϕ ˙ = L =  const.  {:(3.1568)(d)/(ds)[2(1-(r_(**))/(r))c^(2)(t^(˙))]=0quad:}quad=>quad2(1-(r_(**))/(r))c^(2)t^(˙)=E=" const., ",子quad-2r^(2)sin^(2)thetaphi^(˙)=L=" const. "\begin{align*} \frac{d}{d s}\left[2\left(1-\frac{r_{*}}{r}\right) c^{2} \dot{t}\right] & =0 \quad \tag{3.1568} \end{align*} \quad \Rightarrow \quad 2\left(1-\frac{r_{*}}{r}\right) c^{2} \dot{t}=E=\text { const., }, ~ 子 \quad-2 r^{2} \sin ^{2} \theta \dot{\phi}=L=\text { const. }(3.1568)dds[2(1rr)c2t˙]=02(1rr)c2t˙=E= const., , 2r2sin2θϕ˙=L= const. 
Using the initial conditions, we can determine E E EEE and L L LLL as
(3.1570) 2 ( 1 r r 0 ) c 2 β = E , 2 r 0 2 sin 2 θ 0 0 = L (3.1570) 2 1 r r 0 c 2 β = E , 2 r 0 2 sin 2 θ 0 0 = L {:(3.1570)2(1-(r_(**))/(r_(0)))c^(2)beta=E","quad-2r_(0)^(2)sin^(2)theta_(0)*0=L:}\begin{equation*} 2\left(1-\frac{r_{*}}{r_{0}}\right) c^{2} \beta=E, \quad-2 r_{0}^{2} \sin ^{2} \theta_{0} \cdot 0=L \tag{3.1570} \end{equation*}(3.1570)2(1rr0)c2β=E,2r02sin2θ00=L
which means that L = 0 L = 0 L=0L=0L=0, independent of the value of θ 0 θ 0 theta_(0)\theta_{0}θ0 that is not known. Therefore, assuming 0 r r 0 0 r r 0 0 <= r <= r_(0)0 \leq r \leq r_{0}0rr0 and 0 θ π 0 θ π 0 <= theta <= pi0 \leq \theta \leq \pi0θπ, we have that ϕ ˙ = 0 ϕ ˙ = 0 phi^(˙)=0\dot{\phi}=0ϕ˙=0. Furthermore, the Euler-Lagrange equation for the θ θ theta\thetaθ-coordinate yields
L θ = r 2 sin 2 θ ϕ ˙ 2 , d d s L θ = 2 ( 2 r r ˙ θ ˙ + r 2 θ ¨ ) (3.1571) θ ¨ + 2 r r ˙ θ ˙ 1 2 sin 2 θ ϕ ˙ 2 = 0 L θ = r 2 sin 2 θ ϕ ˙ 2 , d d s L θ = 2 2 r r ˙ θ ˙ + r 2 θ ¨ (3.1571) θ ¨ + 2 r r ˙ θ ˙ 1 2 sin 2 θ ϕ ˙ 2 = 0 {:[(delL)/(del theta)=-r^(2)sin 2thetaphi^(˙)^(2)","quad(d)/(ds)(delL)/(del theta)=-2(2r(r^(˙))(theta^(˙))+r^(2)(theta^(¨)))],[(3.1571)=>quadtheta^(¨)+2rr^(˙)theta^(˙)-(1)/(2)sin 2thetaphi^(˙)^(2)=0]:}\begin{gather*} \frac{\partial \mathcal{L}}{\partial \theta}=-r^{2} \sin 2 \theta \dot{\phi}^{2}, \quad \frac{d}{d s} \frac{\partial \mathcal{L}}{\partial \theta}=-2\left(2 r \dot{r} \dot{\theta}+r^{2} \ddot{\theta}\right) \\ \Rightarrow \quad \ddot{\theta}+2 r \dot{r} \dot{\theta}-\frac{1}{2} \sin 2 \theta \dot{\phi}^{2}=0 \tag{3.1571} \end{gather*}Lθ=r2sin2θϕ˙2,ddsLθ=2(2rr˙θ˙+r2θ¨)(3.1571)θ¨+2rr˙θ˙12sin2θϕ˙2=0
Now, since ϕ ˙ = 0 ϕ ˙ = 0 phi^(˙)=0\dot{\phi}=0ϕ˙=0, we also have L θ = r 2 sin 2 θ ϕ ˙ 2 = r 2 sin 2 θ 0 = 0 L θ = r 2 sin 2 θ ϕ ˙ 2 = r 2 sin 2 θ 0 = 0 (delL)/(del theta)=-r^(2)sin 2thetaphi^(˙)^(2)=-r^(2)sin 2theta*0=0\frac{\partial \mathcal{L}}{\partial \theta}=-r^{2} \sin 2 \theta \dot{\phi}^{2}=-r^{2} \sin 2 \theta \cdot 0=0Lθ=r2sin2θϕ˙2=r2sin2θ0=0, which means that we have another constant of motion stemming from the Euler-Lagrange equation for θ θ theta\thetaθ (due to the initial conditions), namely
(3.1572) d d s ( 2 r 2 θ ˙ ) = 0 2 r 2 θ ˙ = A = const. (3.1572) d d s 2 r 2 θ ˙ = 0 2 r 2 θ ˙ = A =  const.  {:(3.1572)(d)/(ds)(-2r^(2)(theta^(˙)))=0quad=>quad-2r^(2)theta^(˙)=A=" const. ":}\begin{equation*} \frac{d}{d s}\left(-2 r^{2} \dot{\theta}\right)=0 \quad \Rightarrow \quad-2 r^{2} \dot{\theta}=A=\text { const. } \tag{3.1572} \end{equation*}(3.1572)dds(2r2θ˙)=02r2θ˙=A= const. 
Again, using the initial conditions, we can determine A A AAA as 2 r 0 2 0 = A 2 r 0 2 0 = A -2r_(0)^(2)*0=A-2 r_{0}^{2} \cdot 0=A2r020=A, which also means that A = 0 A = 0 A=0A=0A=0. Therefore, assuming 0 r r 0 0 r r 0 0 <= r <= r_(0)0 \leq r \leq r_{0}0rr0 as before, we have 2 r 2 θ ˙ = 0 2 r 2 θ ˙ = 0 -2r^(2)theta^(˙)=0-2 r^{2} \dot{\theta}=02r2θ˙=0, which leads to θ ˙ = 0 θ ˙ = 0 theta^(˙)=0\dot{\theta}=0θ˙=0. Thus, based on the initial conditions, the given standard Schwarzschild metric is reduced to
(3.1573) d s 2 = ( 1 r r ) c 2 d t 2 ( 1 r r ) 1 d r 2 (3.1573) d s 2 = 1 r r c 2 d t 2 1 r r 1 d r 2 {:(3.1573)ds^(2)=(1-(r_(**))/(r))c^(2)dt^(2)-(1-(r_(**))/(r))^(-1)dr^(2):}\begin{equation*} d s^{2}=\left(1-\frac{r_{*}}{r}\right) c^{2} d t^{2}-\left(1-\frac{r_{*}}{r}\right)^{-1} d r^{2} \tag{3.1573} \end{equation*}(3.1573)ds2=(1rr)c2dt2(1rr)1dr2
Now, using the constant of motion E E EEE, given by the initial conditions, we find that
(3.1574) i ˙ = E 2 ( 1 r r ) c 2 = ( 1 r r ) 1 E 2 c 2 = d t d s d t = ( 1 r r ) 1 E 2 c 2 d s (3.1574) i ˙ = E 2 1 r r c 2 = 1 r r 1 E 2 c 2 = d t d s d t = 1 r r 1 E 2 c 2 d s {:(3.1574)i^(˙)=(E)/(2(1-(r_(**))/(r))c^(2))=(1-(r_(**))/(r))^(-1)(E)/(2c^(2))=(dt)/(ds)quad=>quad dt=(1-(r_(**))/(r))^(-1)(E)/(2c^(2))ds:}\begin{equation*} \dot{i}=\frac{E}{2\left(1-\frac{r_{*}}{r}\right) c^{2}}=\left(1-\frac{r_{*}}{r}\right)^{-1} \frac{E}{2 c^{2}}=\frac{d t}{d s} \quad \Rightarrow \quad d t=\left(1-\frac{r_{*}}{r}\right)^{-1} \frac{E}{2 c^{2}} d s \tag{3.1574} \end{equation*}(3.1574)i˙=E2(1rr)c2=(1rr)1E2c2=dtdsdt=(1rr)1E2c2ds
Inserting the expression for d t d t dtd tdt into the reduced standard Schwarzschild metric, we obtain
d s 2 = ( 1 r r ) c 2 ( 1 r r ) 2 E 2 4 c 4 ( 1 r r ) 1 d r 2 (3.1575) = ( 1 r r ) 1 E 2 4 c 2 d s 2 ( 1 r r ) 1 d r 2 d s 2 = 1 r r c 2 1 r r 2 E 2 4 c 4 1 r r 1 d r 2 (3.1575) = 1 r r 1 E 2 4 c 2 d s 2 1 r r 1 d r 2 {:[ds^(2)=(1-(r_(**))/(r))c^(2)(1-(r_(**))/(r))^(-2)(E^(2))/(4c^(4))-(1-(r_(**))/(r))^(-1)dr^(2)],[(3.1575)=(1-(r_(**))/(r))^(-1)(E^(2))/(4c^(2))ds^(2)-(1-(r_(**))/(r))^(-1)dr^(2)]:}\begin{align*} d s^{2} & =\left(1-\frac{r_{*}}{r}\right) c^{2}\left(1-\frac{r_{*}}{r}\right)^{-2} \frac{E^{2}}{4 c^{4}}-\left(1-\frac{r_{*}}{r}\right)^{-1} d r^{2} \\ & =\left(1-\frac{r_{*}}{r}\right)^{-1} \frac{E^{2}}{4 c^{2}} d s^{2}-\left(1-\frac{r_{*}}{r}\right)^{-1} d r^{2} \tag{3.1575} \end{align*}ds2=(1rr)c2(1rr)2E24c4(1rr)1dr2(3.1575)=(1rr)1E24c2ds2(1rr)1dr2
which can be written as
(3.1576) d r 2 = [ E 2 4 c 2 ( 1 r r ) ] d s 2 ( d r d s ) = E 2 4 c 2 ( 1 r r ) (3.1576) d r 2 = E 2 4 c 2 1 r r d s 2 d r d s = E 2 4 c 2 1 r r {:(3.1576)dr^(2)=[(E^(2))/(4c^(2))-(1-(r_(**))/(r))]ds^(2)=>((dr)/(ds))=(E^(2))/(4c^(2))-(1-(r_(**))/(r)):}\begin{equation*} d r^{2}=\left[\frac{E^{2}}{4 c^{2}}-\left(1-\frac{r_{*}}{r}\right)\right] d s^{2} \Rightarrow\left(\frac{d r}{d s}\right)=\frac{E^{2}}{4 c^{2}}-\left(1-\frac{r_{*}}{r}\right) \tag{3.1576} \end{equation*}(3.1576)dr2=[E24c2(1rr)]ds2(drds)=E24c2(1rr)
Since the spaceship is freely falling from r 0 r 0 r_(0)r_{0}r0 toward the true singularity at r = 0 r = 0 r=0r=0r=0 in a Schwarzschild black hole, we have r ˙ = d r d s < 0 r ˙ = d r d s < 0 r^(˙)=(dr)/(ds) < 0\dot{r}=\frac{d r}{d s}<0r˙=drds<0, which means that
(3.1577) d r d s = E 2 4 c 2 ( 1 r r ) = d r c d τ d τ = d r c E 2 4 c 2 ( 1 r r ) (3.1577) d r d s = E 2 4 c 2 1 r r = d r c d τ d τ = d r c E 2 4 c 2 1 r r {:(3.1577)(dr)/(ds)=-sqrt((E^(2))/(4c^(2))-(1-(r_(**))/(r)))=(dr)/(cd tau)=>d tau=-(dr)/(csqrt((E^(2))/(4c^(2))-(1-(r_(**))/(r)))):}\begin{equation*} \frac{d r}{d s}=-\sqrt{\frac{E^{2}}{4 c^{2}}-\left(1-\frac{r_{*}}{r}\right)}=\frac{d r}{c d \tau} \Rightarrow d \tau=-\frac{d r}{c \sqrt{\frac{E^{2}}{4 c^{2}}-\left(1-\frac{r_{*}}{r}\right)}} \tag{3.1577} \end{equation*}(3.1577)drds=E24c2(1rr)=drcdτdτ=drcE24c2(1rr)
where τ τ tau\tauτ is the proper time. Integrating the expression for d τ d τ d taud \taudτ from r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0 (at τ = 0 τ = 0 tau=0\tau=0τ=0 ) to r = 0 r = 0 r=0r=0r=0, we find the proper time τ τ tau\tauτ needed to reach the singularity r = 0 r = 0 r=0r=0r=0 (when starting from r = r 0 < r r = r 0 < r r=r_(0) < r_(**)r=r_{0}<r_{*}r=r0<r ) as
(3.1578) τ = r 0 0 d r c E 2 4 c 2 ( 1 r r ) = 0 r 0 d r c E 2 4 c 2 ( 1 r r ) (3.1578) τ = r 0 0 d r c E 2 4 c 2 1 r r = 0 r 0 d r c E 2 4 c 2 1 r r {:(3.1578)tau=-int_(r_(0))^(0)(dr)/(csqrt((E^(2))/(4c^(2))-(1-(r_(**))/(r))))=int_(0)^(r_(0))(dr)/(csqrt((E^(2))/(4c^(2))-(1-(r_(**))/(r)))):}\begin{equation*} \tau=-\int_{r_{0}}^{0} \frac{d r}{c \sqrt{\frac{E^{2}}{4 c^{2}}-\left(1-\frac{r_{*}}{r}\right)}}=\int_{0}^{r_{0}} \frac{d r}{c \sqrt{\frac{E^{2}}{4 c^{2}}-\left(1-\frac{r_{*}}{r}\right)}} \tag{3.1578} \end{equation*}(3.1578)τ=r00drcE24c2(1rr)=0r0drcE24c2(1rr)
However, note that (see above) the constant of motion E E EEE can be written in terms of the initial conditions as E = 2 ( 1 r r 0 ) c 2 β E = 2 1 r r 0 c 2 β E=2(1-(r_(**))/(r_(0)))c^(2)betaE=2\left(1-\frac{r_{*}}{r_{0}}\right) c^{2} \betaE=2(1rr0)c2β, which means that the proper time τ τ tau\tauτ is given by
(3.1579) τ = 0 r 0 d r c ( 1 r s r 0 ) 2 c 2 β 2 ( 1 r r ) 0 r 0 f ( r ) d r (3.1579) τ = 0 r 0 d r c 1 r s r 0 2 c 2 β 2 1 r r 0 r 0 f ( r ) d r {:(3.1579)tau=int_(0)^(r_(0))(dr)/(csqrt((1-(r_(s))/(r_(0)))^(2)c^(2)beta^(2)-(1-(r_(**))/(r))))-=int_(0)^(r_(0))f(r)dr:}\begin{equation*} \tau=\int_{0}^{r_{0}} \frac{d r}{c \sqrt{\left(1-\frac{r_{s}}{r_{0}}\right)^{2} c^{2} \beta^{2}-\left(1-\frac{r_{*}}{r}\right)}} \equiv \int_{0}^{r_{0}} f(r) d r \tag{3.1579} \end{equation*}(3.1579)τ=0r0drc(1rsr0)2c2β2(1rr)0r0f(r)dr
Thus, we can finally identify the function f ( r ) f ( r ) f(r)f(r)f(r) as
(3.1580) f ( r ) = 1 c ( 1 r r 0 ) 2 c 2 β 2 ( 1 r r ) (3.1580) f ( r ) = 1 c 1 r r 0 2 c 2 β 2 1 r r {:(3.1580)f(r)=(1)/(csqrt((1-(r_(**))/(r_(0)))^(2)c^(2)beta^(2)-(1-(r_(**))/(r)))):}\begin{equation*} f(r)=\frac{1}{c \sqrt{\left(1-\frac{r_{*}}{r_{0}}\right)^{2} c^{2} \beta^{2}-\left(1-\frac{r_{*}}{r}\right)}} \tag{3.1580} \end{equation*}(3.1580)f(r)=1c(1rr0)2c2β2(1rr)
2.92
Write the geodesic equations in ( x 0 , r , θ , ϕ ) x 0 , r , θ , ϕ (x^(0),r,theta,phi)\left(x^{0}, r, \theta, \phi\right)(x0,r,θ,ϕ) coordinates in the plane θ = π 2 θ = π 2 theta=(pi)/(2)\theta=\frac{\pi}{2}θ=π2, namely
(3.1581) x ¨ 0 + α r 2 ( 1 α r ) 1 x ˙ 0 r ˙ = 0 (3.1582) r ¨ + α 2 r 2 ( 1 α r ) ( x ˙ 0 ) 2 α 2 r 2 ( 1 α r ) 1 r ˙ 2 r ( 1 α r ) ϕ ˙ 2 = 0 (3.1583) ϕ ¨ + 2 r r ˙ ϕ ˙ = 0 (3.1581) x ¨ 0 + α r 2 1 α r 1 x ˙ 0 r ˙ = 0 (3.1582) r ¨ + α 2 r 2 1 α r x ˙ 0 2 α 2 r 2 1 α r 1 r ˙ 2 r 1 α r ϕ ˙ 2 = 0 (3.1583) ϕ ¨ + 2 r r ˙ ϕ ˙ = 0 {:[(3.1581)x^(¨)^(0)+(alpha)/(r^(2))(1-(alpha )/(r))^(-1)x^(˙)^(0)r^(˙)=0],[(3.1582)r^(¨)+(alpha)/(2r^(2))(1-(alpha )/(r))(x^(˙)^(0))^(2)-(alpha)/(2r^(2))(1-(alpha )/(r))^(-1)r^(˙)^(2)-r(1-(alpha )/(r))phi^(˙)^(2)=0],[(3.1583)phi^(¨)+(2)/(r)r^(˙)phi^(˙)=0]:}\begin{align*} & \ddot{x}^{0}+\frac{\alpha}{r^{2}}\left(1-\frac{\alpha}{r}\right)^{-1} \dot{x}^{0} \dot{r}=0 \tag{3.1581}\\ & \ddot{r}+\frac{\alpha}{2 r^{2}}\left(1-\frac{\alpha}{r}\right)\left(\dot{x}^{0}\right)^{2}-\frac{\alpha}{2 r^{2}}\left(1-\frac{\alpha}{r}\right)^{-1} \dot{r}^{2}-r\left(1-\frac{\alpha}{r}\right) \dot{\phi}^{2}=0 \tag{3.1582}\\ & \ddot{\phi}+\frac{2}{r} \dot{r} \dot{\phi}=0 \tag{3.1583} \end{align*}(3.1581)x¨0+αr2(1αr)1x˙0r˙=0(3.1582)r¨+α2r2(1αr)(x˙0)2α2r2(1αr)1r˙2r(1αr)ϕ˙2=0(3.1583)ϕ¨+2rr˙ϕ˙=0
where α = 2 G M c 2 α = 2 G M c 2 alpha=(2GM)/(c^(2))\alpha=\frac{2 G M}{c^{2}}α=2GMc2. For a derivation of the geodesic equations, see Problem 2.70. The geodesic equation for x 0 x 0 x^(0)x^{0}x0 can be integrated at once, which yields
(3.1584) x ˙ 0 = k 1 α r , where k is a constant. (3.1584) x ˙ 0 = k 1 α r ,  where  k  is a constant.  {:(3.1584)x^(˙)^(0)=(k)/(1-(alpha )/(r))","quad" where "k" is a constant. ":}\begin{equation*} \dot{x}^{0}=\frac{k}{1-\frac{\alpha}{r}}, \quad \text { where } k \text { is a constant. } \tag{3.1584} \end{equation*}(3.1584)x˙0=k1αr, where k is a constant. 
On a geodesic for a timelike object, it holds that
(3.1585) 1 = g 00 ( x ˙ 0 ) 2 + g r r r ˙ 2 (3.1585) 1 = g 00 x ˙ 0 2 + g r r r ˙ 2 {:(3.1585)1=g_(00)(x^(˙)^(0))^(2)+g_(rr)r^(˙)^(2):}\begin{equation*} 1=g_{00}\left(\dot{x}^{0}\right)^{2}+g_{r r} \dot{r}^{2} \tag{3.1585} \end{equation*}(3.1585)1=g00(x˙0)2+grrr˙2
where g 00 = 1 α r g 00 = 1 α r g_(00)=1-(alpha )/(r)g_{00}=1-\frac{\alpha}{r}g00=1αr and g r r = ( 1 α r ) 1 g r r = 1 α r 1 g_(rr)=-(1-(alpha )/(r))^(-1)g_{r r}=-\left(1-\frac{\alpha}{r}\right)^{-1}grr=(1αr)1. This implies that
(3.1586) r ˙ = ϵ + α r = d r d s = 1 c d r d τ (3.1586) r ˙ = ϵ + α r = d r d s = 1 c d r d τ {:(3.1586)r^(˙)=-sqrt(epsilon+(alpha )/(r))=(dr)/(ds)=(1)/(c)(dr)/(d tau):}\begin{equation*} \dot{r}=-\sqrt{\epsilon+\frac{\alpha}{r}}=\frac{d r}{d s}=\frac{1}{c} \frac{d r}{d \tau} \tag{3.1586} \end{equation*}(3.1586)r˙=ϵ+αr=drds=1cdrdτ
where ϵ = k 2 1 ϵ = k 2 1 epsilon=k^(2)-1\epsilon=k^{2}-1ϵ=k21. Note the minus sign (which is due to the fact that r r rrr is decreasing, i.e., the observer is freely falling toward the center of the black hole, and hence, r ˙ < 0 r ˙ < 0 r^(˙) < 0\dot{r}<0r˙<0 ). So, we have
(3.1587) d τ d r = 1 c ϵ + α r (3.1587) d τ d r = 1 c ϵ + α r {:(3.1587)(d tau)/(dr)=-(1)/(csqrt(epsilon+(alpha )/(r))):}\begin{equation*} \frac{d \tau}{d r}=-\frac{1}{c \sqrt{\epsilon+\frac{\alpha}{r}}} \tag{3.1587} \end{equation*}(3.1587)dτdr=1cϵ+αr
According to the hint, integration gives
(3.1588) d r ϵ + α r = 1 ϵ y 2 β 2 4 1 ϵ β 2 ln ( y + y 2 β 2 4 ) (3.1588) d r ϵ + α r = 1 ϵ y 2 β 2 4 1 ϵ β 2 ln y + y 2 β 2 4 {:(3.1588)int(dr)/(sqrt(epsilon+(alpha )/(r)))=(1)/(sqrtepsilon)sqrt(y^(2)-(beta^(2))/(4))-(1)/(sqrtepsilon)(beta)/(2)ln(y+sqrt(y^(2)-(beta^(2))/(4))):}\begin{equation*} \int \frac{d r}{\sqrt{\epsilon+\frac{\alpha}{r}}}=\frac{1}{\sqrt{\epsilon}} \sqrt{y^{2}-\frac{\beta^{2}}{4}}-\frac{1}{\sqrt{\epsilon}} \frac{\beta}{2} \ln \left(y+\sqrt{y^{2}-\frac{\beta^{2}}{4}}\right) \tag{3.1588} \end{equation*}(3.1588)drϵ+αr=1ϵy2β241ϵβ2ln(y+y2β24)
where β = α ϵ β = α ϵ beta=(alpha )/(epsilon)\beta=\frac{\alpha}{\epsilon}β=αϵ and y = r + β 2 y = r + β 2 y=r+(beta)/(2)y=r+\frac{\beta}{2}y=r+β2. Thus, the (proper) time needed for the interval r 0 r 0 r_(0) <=r_{0} \leqr0 r r 1 r r 1 r <= r_(1)r \leq r_{1}rr1 is given by
(3.1589) Δ τ τ 1 τ 0 = 1 c r 0 r 1 d r ϵ + α r = 1 c r 0 r 1 d r k 2 ( 1 α r ) (3.1589) Δ τ τ 1 τ 0 = 1 c r 0 r 1 d r ϵ + α r = 1 c r 0 r 1 d r k 2 1 α r {:(3.1589)Delta tau-=tau_(1)-tau_(0)=-(1)/(c)int_(r_(0))^(r_(1))(dr)/(sqrt(epsilon+(alpha )/(r)))=-(1)/(c)int_(r_(0))^(r_(1))(dr)/(sqrt(k^(2)-(1-(alpha )/(r)))):}\begin{equation*} \Delta \tau \equiv \tau_{1}-\tau_{0}=-\frac{1}{c} \int_{r_{0}}^{r_{1}} \frac{d r}{\sqrt{\epsilon+\frac{\alpha}{r}}}=-\frac{1}{c} \int_{r_{0}}^{r_{1}} \frac{d r}{\sqrt{k^{2}-\left(1-\frac{\alpha}{r}\right)}} \tag{3.1589} \end{equation*}(3.1589)Δττ1τ0=1cr0r1drϵ+αr=1cr0r1drk2(1αr)
where r 0 = r ( τ 0 ) = 10 10 km r 0 = r τ 0 = 10 10 km r_(0)=r(tau_(0))=10^(10)kmr_{0}=r\left(\tau_{0}\right)=10^{10} \mathrm{~km}r0=r(τ0)=1010 km and r 1 = r ( τ 1 ) = α r 1 = r τ 1 = α r_(1)=r(tau_(1))=alphar_{1}=r\left(\tau_{1}\right)=\alphar1=r(τ1)=α, i.e., the Schwarzschild horizon. Now, use the initial condition to calculate ϵ ϵ epsilon\epsilonϵ :
(3.1590) v 0 = c d r d x 0 = c r ˙ x ˙ 0 at τ = τ 0 (3.1590) v 0 = c d r d x 0 = c r ˙ x ˙ 0  at  τ = τ 0 {:(3.1590)-v_(0)=c(dr)/(dx^(0))=c((r^(˙)))/(x^(˙)^(0))quad" at "tau=tau_(0):}\begin{equation*} -v_{0}=c \frac{d r}{d x^{0}}=c \frac{\dot{r}}{\dot{x}^{0}} \quad \text { at } \tau=\tau_{0} \tag{3.1590} \end{equation*}(3.1590)v0=cdrdx0=cr˙x˙0 at τ=τ0
which implies that
(3.1591) v 0 = c k ϵ + α r 0 ( 1 α r 0 ) k 2 = ( 1 α r 0 ) [ 1 v 0 2 c 2 ( 1 α r 0 ) 2 ] 1 (3.1591) v 0 = c k ϵ + α r 0 1 α r 0 k 2 = 1 α r 0 1 v 0 2 c 2 1 α r 0 2 1 {:(3.1591)v_(0)=(c)/(k)sqrt(epsilon+(alpha)/(r_(0)))(1-(alpha)/(r_(0)))=>k^(2)=(1-(alpha)/(r_(0)))[1-(v_(0)^(2))/(c^(2))(1-(alpha)/(r_(0)))^(-2)]^(-1):}\begin{equation*} v_{0}=\frac{c}{k} \sqrt{\epsilon+\frac{\alpha}{r_{0}}}\left(1-\frac{\alpha}{r_{0}}\right) \Rightarrow k^{2}=\left(1-\frac{\alpha}{r_{0}}\right)\left[1-\frac{v_{0}^{2}}{c^{2}}\left(1-\frac{\alpha}{r_{0}}\right)^{-2}\right]^{-1} \tag{3.1591} \end{equation*}(3.1591)v0=ckϵ+αr0(1αr0)k2=(1αr0)[1v02c2(1αr0)2]1
Inserting k 2 k 2 k^(2)k^{2}k2 into the (proper) time interval Δ τ Δ τ Delta tau\Delta \tauΔτ, we obtain
(3.1592) Δ τ = 1 c r 0 r 1 d r ( 1 α r 0 ) [ 1 v 0 2 c 2 ( 1 α r 0 ) 2 ] 1 ( 1 α r ) (3.1592) Δ τ = 1 c r 0 r 1 d r 1 α r 0 1 v 0 2 c 2 1 α r 0 2 1 1 α r {:(3.1592)Delta tau=-(1)/(c)int_(r_(0))^(r_(1))(dr)/(sqrt((1-(alpha)/(r_(0)))[1-(v_(0)^(2))/(c^(2))(1-(alpha)/(r_(0)))^(-2)]^(-1)-(1-(alpha )/(r)))):}\begin{equation*} \Delta \tau=-\frac{1}{c} \int_{r_{0}}^{r_{1}} \frac{d r}{\sqrt{\left(1-\frac{\alpha}{r_{0}}\right)\left[1-\frac{v_{0}^{2}}{c^{2}}\left(1-\frac{\alpha}{r_{0}}\right)^{-2}\right]^{-1}-\left(1-\frac{\alpha}{r}\right)}} \tag{3.1592} \end{equation*}(3.1592)Δτ=1cr0r1dr(1αr0)[1v02c2(1αr0)2]1(1αr)
which can be computed analytically using the hint, but the result is too lengthy for it to be useful to display. Computing the integral numerically, the result is Δ τ Δ τ Delta tau≃\Delta \tau \simeqΔτ 5.85615 10 8 s 5.85615 10 8 s 5.85615*10^(8)s5.85615 \cdot 10^{8} \mathrm{~s}5.85615108 s, which corresponds to about 18.6 years. Thus, the proper time needed for the observer to reach the Schwarzschild horizon is about 18.6 years.
However, the quantities α , r 0 α , r 0 alpha,r_(0)\alpha, r_{0}α,r0, and v 0 v 0 v_(0)v_{0}v0 are given, so the constant ϵ = k 2 1 ϵ = k 2 1 epsilon=k^(2)-1\epsilon=k^{2}-1ϵ=k21 can be uniquely computed from them. Assuming r 0 α r 0 α r_(0)≫alphar_{0} \gg \alphar0α, qualitative estimates of the parameters are as follows
v 0 c k ϵ , ϵ = k 2 1 v 0 2 c 2 k 2 , k 1 1 v 0 2 / c 2 ( 1 ) , (3.1593) ϵ = k 2 1 v 0 2 c 2 v 0 2 ( v 0 2 c 2 ) , and β 2 G M v 0 2 ( 1 v 0 2 c 2 ) ( 2 G M v 0 2 ) , v 0 c k ϵ , ϵ = k 2 1 v 0 2 c 2 k 2 , k 1 1 v 0 2 / c 2 ( 1 ) , (3.1593) ϵ = k 2 1 v 0 2 c 2 v 0 2 v 0 2 c 2 ,  and  β 2 G M v 0 2 1 v 0 2 c 2 2 G M v 0 2 , {:[v_(0)≃(c)/(k)sqrtepsilon","quad epsilon=k^(2)-1≃(v_(0)^(2))/(c^(2))*k^(2)","quad k≃(1)/(sqrt(1-v_(0)^(2)//c^(2)))(∼1)","],[(3.1593)epsilon=k^(2)-1≃(v_(0)^(2))/(c^(2)-v_(0)^(2))(∼(v_(0)^(2))/(c^(2)))","quad" and "quad beta≃(2GM)/(v_(0)^(2))(1-(v_(0)^(2))/(c^(2)))(∼(2GM)/(v_(0)^(2)))","]:}\begin{gather*} v_{0} \simeq \frac{c}{k} \sqrt{\epsilon}, \quad \epsilon=k^{2}-1 \simeq \frac{v_{0}^{2}}{c^{2}} \cdot k^{2}, \quad k \simeq \frac{1}{\sqrt{1-v_{0}^{2} / c^{2}}}(\sim 1), \\ \epsilon=k^{2}-1 \simeq \frac{v_{0}^{2}}{c^{2}-v_{0}^{2}}\left(\sim \frac{v_{0}^{2}}{c^{2}}\right), \quad \text { and } \quad \beta \simeq \frac{2 G M}{v_{0}^{2}}\left(1-\frac{v_{0}^{2}}{c^{2}}\right)\left(\sim \frac{2 G M}{v_{0}^{2}}\right), \tag{3.1593} \end{gather*}v0ckϵ,ϵ=k21v02c2k2,k11v02/c2(1),(3.1593)ϵ=k21v02c2v02(v02c2), and β2GMv02(1v02c2)(2GMv02),
where limits within parentheses also hold if it is assumed that v 0 c v 0 c v_(0)≪cv_{0} \ll cv0c. Therefore, using the estimates of the parameters, we can approximate the integral for the (proper) time interval Δ τ Δ τ Delta tau\Delta \tauΔτ as
Δ τ 1 v 0 r 0 r 1 d r 1 + 2 G M v 0 2 r r 0 v 0 ( 1 + 2 G M v 0 2 r 0 2 G M v 0 2 r 0 arsinh v 0 2 r 0 2 G M ) (3.1594) 4.32267 10 8 s 13.7 years Δ τ 1 v 0 r 0 r 1 d r 1 + 2 G M v 0 2 r r 0 v 0 1 + 2 G M v 0 2 r 0 2 G M v 0 2 r 0 arsinh v 0 2 r 0 2 G M (3.1594) 4.32267 10 8 s 13.7  years  {:[Delta tau≃-(1)/(v_(0))int_(r_(0))^(r_(1))(dr)/(sqrt(1+(2GM)/(v_(0)^(2)r)))≃(r_(0))/(v_(0))(sqrt(1+(2GM)/(v_(0)^(2)r_(0)))-(2GM)/(v_(0)^(2)r_(0))*arsinhsqrt((v_(0)^(2)r_(0))/(2GM)))],[(3.1594)≃4.32267*10^(8)s∼13.7" years "]:}\begin{align*} \Delta \tau & \simeq-\frac{1}{v_{0}} \int_{r_{0}}^{r_{1}} \frac{d r}{\sqrt{1+\frac{2 G M}{v_{0}^{2} r}}} \simeq \frac{r_{0}}{v_{0}}\left(\sqrt{1+\frac{2 G M}{v_{0}^{2} r_{0}}}-\frac{2 G M}{v_{0}^{2} r_{0}} \cdot \operatorname{arsinh} \sqrt{\frac{v_{0}^{2} r_{0}}{2 G M}}\right) \\ & \simeq 4.32267 \cdot 10^{8} \mathrm{~s} \sim 13.7 \text { years } \tag{3.1594} \end{align*}Δτ1v0r0r1dr1+2GMv02rr0v0(1+2GMv02r02GMv02r0arsinhv02r02GM)(3.1594)4.32267108 s13.7 years 
which means that the approximation underestimates (by about 25 % 25 % 25%25 \%25% ) the true value of the proper time, but it gives the correct order of magnitude.

2.93

Due to the Schwarzschild metric components not depending on t t ttt, we can conclude that t t del_(t)\partial_{t}t is a Killing vector field. It follows that a freely falling particle with a worldline that is a geodesic has a constant of motion
(3.1595) k = g ( t , γ ˙ ) = g 00 x ˙ 0 (3.1595) k = g t , γ ˙ = g 00 x ˙ 0 {:(3.1595)k=g(del_(t),(gamma^(˙)))=g_(00)x^(˙)^(0):}\begin{equation*} k=g\left(\partial_{t}, \dot{\gamma}\right)=g_{00} \dot{x}^{0} \tag{3.1595} \end{equation*}(3.1595)k=g(t,γ˙)=g00x˙0
Since m m mmm and c c ccc are constants, it also follows that p 0 = m c k p 0 = m c k p_(0)=mckp_{0}=m c kp0=mck is a constant. Furthermore, the normalization of the 4 -velocity implies that
(3.1596) L = g μ ν x ˙ μ x ˙ v = g 00 c 2 t ˙ 2 g r r r ˙ 2 = 1 , (3.1596) L = g μ ν x ˙ μ x ˙ v = g 00 c 2 t ˙ 2 g r r r ˙ 2 = 1 , {:(3.1596)L=g_(mu nu)x^(˙)^(mu)x^(˙)^(v)=g_(00)c^(2)t^(˙)^(2)-g_(rr)r^(˙)^(2)=1",":}\begin{equation*} \mathcal{L}=g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{v}=g_{00} c^{2} \dot{t}^{2}-g_{r r} \dot{r}^{2}=1, \tag{3.1596} \end{equation*}(3.1596)L=gμνx˙μx˙v=g00c2t˙2grrr˙2=1,
for a particle falling radially. With E = p 0 / c E = p 0 / c E=p_(0)//cE=p_{0} / cE=p0/c, this implies that
1 = g 00 ( x ˙ 0 ) 2 + g r r r ˙ 2 = 1 g 00 [ E 2 ( m c 2 ) 2 r ˙ 2 ] (3.1597) r ˙ = ( E m c 2 ) 2 ( 1 r r ) 1 = g 00 x ˙ 0 2 + g r r r ˙ 2 = 1 g 00 E 2 m c 2 2 r ˙ 2 (3.1597) r ˙ = E m c 2 2 1 r r {:[1=g_(00)(x^(˙)^(0))^(2)+g_(rr)r^(˙)^(2)=(1)/(g_(00))*[(E^(2))/((mc^(2))^(2))-r^(˙)^(2)]],[(3.1597) Longrightarrowquadr^(˙)=-sqrt(((E)/(mc^(2)))^(2)-(1-(r_(**))/(r)))]:}\begin{align*} 1= & g_{00}\left(\dot{x}^{0}\right)^{2}+g_{r r} \dot{r}^{2}=\frac{1}{g_{00}} \cdot\left[\frac{E^{2}}{\left(m c^{2}\right)^{2}}-\dot{r}^{2}\right] \\ & \Longrightarrow \quad \dot{r}=-\sqrt{\left(\frac{E}{m c^{2}}\right)^{2}-\left(1-\frac{r_{*}}{r}\right)} \tag{3.1597} \end{align*}1=g00(x˙0)2+grrr˙2=1g00[E2(mc2)2r˙2](3.1597)r˙=(Emc2)2(1rr)
with the negative sign being due to r r rrr decreasing with proper time. Integrating from r = 3 r / 2 r = 3 r / 2 r=3r_(**)//2r=3 r_{*} / 2r=3r/2 to r = r r = r r=r_(**)r=r_{*}r=r, we find that
(3.1598) Δ s = 3 r / r r d r ( E m c 2 ) 2 ( 1 r r ) = r 3 r / r d r ( E m c 2 ) 2 ( 1 r r ) (3.1598) Δ s = 3 r / r r d r E m c 2 2 1 r r = r 3 r / r d r E m c 2 2 1 r r {:(3.1598)Delta s=-int_(3r_(**)//r)^(r_(**))(dr)/(sqrt(((E)/(mc^(2)))^(2)-(1-(r_(**))/(r))))=int_(r_(**))^(3r_(**)//r)(dr)/(sqrt(((E)/(mc^(2)))^(2)-(1-(r_(**))/(r)))):}\begin{equation*} \Delta s=-\int_{3 r_{*} / r}^{r_{*}} \frac{d r}{\sqrt{\left(\frac{E}{m c^{2}}\right)^{2}-\left(1-\frac{r_{*}}{r}\right)}}=\int_{r_{*}}^{3 r_{*} / r} \frac{d r}{\sqrt{\left(\frac{E}{m c^{2}}\right)^{2}-\left(1-\frac{r_{*}}{r}\right)}} \tag{3.1598} \end{equation*}(3.1598)Δs=3r/rrdr(Emc2)2(1rr)=r3r/rdr(Emc2)2(1rr)

2.94

The worldline of the observer is given by
(3.1599) t = β τ , r = r 0 , θ = π 2 , φ = ω τ (3.1599) t = β τ , r = r 0 , θ = π 2 , φ = ω τ {:(3.1599)t=beta tau","quad r=r_(0)","quad theta=(pi)/(2)","quad varphi=omega tau:}\begin{equation*} t=\beta \tau, \quad r=r_{0}, \quad \theta=\frac{\pi}{2}, \quad \varphi=\omega \tau \tag{3.1599} \end{equation*}(3.1599)t=βτ,r=r0,θ=π2,φ=ωτ
where τ τ tau\tauτ is the proper time. To find the constant β β beta\betaβ, we note that normalization of the 4 -velocity yields
(3.1600) g t t β 2 + g φ φ 2 ω 2 = 1 , (3.1600) g t t β 2 + g φ φ 2 ω 2 = 1 , {:(3.1600)g_(tt)beta^(2)+g_(varphi varphi)^(2)omega^(2)=1",":}\begin{equation*} g_{t t} \beta^{2}+g_{\varphi \varphi}^{2} \omega^{2}=1, \tag{3.1600} \end{equation*}(3.1600)gttβ2+gφφ2ω2=1,
which implies
(3.1601) β = 1 + r 0 2 ω 2 1 r / r 0 (3.1601) β = 1 + r 0 2 ω 2 1 r / r 0 {:(3.1601)beta=sqrt((1+r_(0)^(2)omega^(2))/(1-r_(**)//r_(0))):}\begin{equation*} \beta=\sqrt{\frac{1+r_{0}^{2} \omega^{2}}{1-r_{*} / r_{0}}} \tag{3.1601} \end{equation*}(3.1601)β=1+r02ω21r/r0
where r = 2 G M r = 2 G M r_(**)=2GMr_{*}=2 G Mr=2GM is the Schwarzschild radius. With the 4 -velocity being V = β t + ω φ V = β t + ω φ V=betadel_(t)+omegadel_(varphi)V=\beta \partial_{t}+\omega \partial_{\varphi}V=βt+ωφ, we generally find that
A = V V = β 2 t t + β ω ( t φ + φ t ) + ω 2 φ φ (3.1602) = ( β 2 Γ t t a + 2 β ω Γ t φ a + ω 2 Γ φ φ a ) a A = V V = β 2 t t + β ω t φ + φ t + ω 2 φ φ (3.1602) = β 2 Γ t t a + 2 β ω Γ t φ a + ω 2 Γ φ φ a a {:[A=grad_(V)V=beta^(2)grad_(t)del_(t)+beta omega(grad_(t)del_(varphi)+grad_(varphi)del_(t))+omega^(2)grad_(varphi)del_(varphi)],[(3.1602)=(beta^(2)Gamma_(tt)^(a)+2beta omegaGamma_(t varphi)^(a)+omega^(2)Gamma_(varphi varphi)^(a))del_(a)]:}\begin{align*} A=\nabla_{V} V & =\beta^{2} \nabla_{t} \partial_{t}+\beta \omega\left(\nabla_{t} \partial_{\varphi}+\nabla_{\varphi} \partial_{t}\right)+\omega^{2} \nabla_{\varphi} \partial_{\varphi} \\ & =\left(\beta^{2} \Gamma_{t t}^{a}+2 \beta \omega \Gamma_{t \varphi}^{a}+\omega^{2} \Gamma_{\varphi \varphi}^{a}\right) \partial_{a} \tag{3.1602} \end{align*}A=VV=β2tt+βω(tφ+φt)+ω2φφ(3.1602)=(β2Γtta+2βωΓtφa+ω2Γφφa)a
The relevant nonzero Christoffel symbols of the Schwarzschild metric are
(3.1603) Γ t t r = r ( r r ) 2 r 3 , Γ φ φ r = ( r r ) sin 2 ( θ ) , Γ φ φ θ = sin ( θ ) cos ( θ ) (3.1603) Γ t t r = r r r 2 r 3 , Γ φ φ r = r r sin 2 ( θ ) , Γ φ φ θ = sin ( θ ) cos ( θ ) {:(3.1603)Gamma_(tt)^(r)=(r_(**)(r-r_(**)))/(2r^(3))","quadGamma_(varphi varphi)^(r)=-(r-r_(**))sin^(2)(theta)","quadGamma_(varphi varphi)^(theta)=-sin(theta)cos(theta):}\begin{equation*} \Gamma_{t t}^{r}=\frac{r_{*}\left(r-r_{*}\right)}{2 r^{3}}, \quad \Gamma_{\varphi \varphi}^{r}=-\left(r-r_{*}\right) \sin ^{2}(\theta), \quad \Gamma_{\varphi \varphi}^{\theta}=-\sin (\theta) \cos (\theta) \tag{3.1603} \end{equation*}(3.1603)Γttr=r(rr)2r3,Γφφr=(rr)sin2(θ),Γφφθ=sin(θ)cos(θ)
However, θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2 leads to sin ( θ ) = 1 sin ( θ ) = 1 sin(theta)=1\sin (\theta)=1sin(θ)=1 and cos ( θ ) = 0 cos ( θ ) = 0 cos(theta)=0\cos (\theta)=0cos(θ)=0, resulting in
(3.1604) A = [ r 2 ( r 0 2 + 3 ω 2 ) r 0 ω 2 ] r (3.1604) A = r 2 r 0 2 + 3 ω 2 r 0 ω 2 r {:(3.1604)A=[(r_(**))/(2)(r_(0)^(-2)+3omega^(2))-r_(0)omega^(2)]del_(r):}\begin{equation*} A=\left[\frac{r_{*}}{2}\left(r_{0}^{-2}+3 \omega^{2}\right)-r_{0} \omega^{2}\right] \partial_{r} \tag{3.1604} \end{equation*}(3.1604)A=[r2(r02+3ω2)r0ω2]r
For the proper acceleration α α alpha\alphaα, we use that α 2 = g ( A , A ) α 2 = g ( A , A ) alpha^(2)=-g(A,A)\alpha^{2}=-g(A, A)α2=g(A,A), and therefore,
(3.1605) α 2 = r 0 r 0 r [ r 2 ( r 0 2 + 3 ω 2 ) r 0 ω 2 ] 2 (3.1605) α 2 = r 0 r 0 r r 2 r 0 2 + 3 ω 2 r 0 ω 2 2 {:(3.1605)alpha^(2)=(r_(0))/(r_(0)-r_(**))[(r_(**))/(2)*(r_(0)^(-2)+3omega^(2))-r_(0)omega^(2)]^(2):}\begin{equation*} \alpha^{2}=\frac{r_{0}}{r_{0}-r_{*}}\left[\frac{r_{*}}{2} \cdot\left(r_{0}^{-2}+3 \omega^{2}\right)-r_{0} \omega^{2}\right]^{2} \tag{3.1605} \end{equation*}(3.1605)α2=r0r0r[r2(r02+3ω2)r0ω2]2
and consequently,
(3.1606) α = r 0 r 0 r | r 2 ( r 0 2 + 3 ω 2 ) r 0 ω 2 | . (3.1606) α = r 0 r 0 r r 2 r 0 2 + 3 ω 2 r 0 ω 2 . {:(3.1606)alpha=sqrt((r_(0))/(r_(0)-r_(**)))|(r_(**))/(2)(r_(0)^(-2)+3omega^(2))-r_(0)omega^(2)|.:}\begin{equation*} \alpha=\sqrt{\frac{r_{0}}{r_{0}-r_{*}}}\left|\frac{r_{*}}{2}\left(r_{0}^{-2}+3 \omega^{2}\right)-r_{0} \omega^{2}\right| . \tag{3.1606} \end{equation*}(3.1606)α=r0r0r|r2(r02+3ω2)r0ω2|.
Note that, for r r 0 r r 0 r_(**)≪r_(0)r_{*} \ll r_{0}rr0, we find that α = 0 α = 0 alpha=0\alpha=0α=0 when
(3.1607) G M r 0 2 = r 0 ω 2 , (3.1607) G M r 0 2 = r 0 ω 2 , {:(3.1607)(GM)/(r_(0)^(2))=r_(0)omega^(2)",":}\begin{equation*} \frac{G M}{r_{0}^{2}}=r_{0} \omega^{2}, \tag{3.1607} \end{equation*}(3.1607)GMr02=r0ω2,
i.e., when the classical gravitational acceleration is the same as the centripetal acceleration required to keep the orbit at r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0.
2.95
Using the Schwarzschild metric, we obtain the Lagrangian
(3.1608) L = ( 1 r r ) t ˙ 2 ( 1 r r ) 1 r ˙ 2 (3.1608) L = 1 r r t ˙ 2 1 r r 1 r ˙ 2 {:(3.1608)L=(1-(r_(**))/(r))t^(˙)^(2)-(1-(r_(**))/(r))^(-1)r^(˙)^(2):}\begin{equation*} \mathcal{L}=\left(1-\frac{r_{*}}{r}\right) \dot{t}^{2}-\left(1-\frac{r_{*}}{r}\right)^{-1} \dot{r}^{2} \tag{3.1608} \end{equation*}(3.1608)L=(1rr)t˙2(1rr)1r˙2
where r = 2 G M r = 2 G M r_(**)=2GMr_{*}=2 G Mr=2GM is the Schwarzschild radius and the dot means differentiation with respect to proper time τ τ tau\tauτ. There exists a conservation law L = 1 L = 1 L=1\mathcal{L}=1L=1 (remember that we set c = 1 c = 1 c=1c=1c=1 ) and we have the Euler-Lagrange equation
(3.1609) d d τ L t ˙ L t = d d τ [ 2 ( 1 r r ) t ˙ ] = 0 i ˙ = C ( 1 r r ) 1 (3.1609) d d τ L t ˙ L t = d d τ 2 1 r r t ˙ = 0 i ˙ = C 1 r r 1 {:(3.1609)(d)/(d tau)(delL)/(del(t^(˙)))-(delL)/(del t)=(d)/(d tau)[2(1-(r_(**))/(r))(t^(˙))]=0quad=>quadi^(˙)=C(1-(r_(**))/(r))^(-1):}\begin{equation*} \frac{d}{d \tau} \frac{\partial \mathcal{L}}{\partial \dot{t}}-\frac{\partial \mathcal{L}}{\partial t}=\frac{d}{d \tau}\left[2\left(1-\frac{r_{*}}{r}\right) \dot{t}\right]=0 \quad \Rightarrow \quad \dot{i}=C\left(1-\frac{r_{*}}{r}\right)^{-1} \tag{3.1609} \end{equation*}(3.1609)ddτLt˙Lt=ddτ[2(1rr)t˙]=0i˙=C(1rr)1
for some integration constant C C CCC. Combining the conservation law and the EulerLagrange equation yields
L 1 = ( 1 r r ) C 2 ( 1 r r ) 2 ( 1 r r ) 1 r ˙ 2 1 (3.1610) = ( 1 r r ) 1 ( C 2 r ˙ 2 ) 1 = 0 L 1 = 1 r r C 2 1 r r 2 1 r r 1 r ˙ 2 1 (3.1610) = 1 r r 1 C 2 r ˙ 2 1 = 0 {:[L-1=(1-(r_(**))/(r))C^(2)(1-(r_(**))/(r))^(-2)-(1-(r_(**))/(r))^(-1)r^(˙)^(2)-1],[(3.1610)=(1-(r_(**))/(r))^(-1)(C^(2)-r^(˙)^(2))-1=0]:}\begin{align*} \mathcal{L}-1 & =\left(1-\frac{r_{*}}{r}\right) C^{2}\left(1-\frac{r_{*}}{r}\right)^{-2}-\left(1-\frac{r_{*}}{r}\right)^{-1} \dot{r}^{2}-1 \\ & =\left(1-\frac{r_{*}}{r}\right)^{-1}\left(C^{2}-\dot{r}^{2}\right)-1=0 \tag{3.1610} \end{align*}L1=(1rr)C2(1rr)2(1rr)1r˙21(3.1610)=(1rr)1(C2r˙2)1=0
i.e.,
(3.1611) r ˙ 2 = C 2 ( 1 r r ) (3.1611) r ˙ 2 = C 2 1 r r {:(3.1611)r^(˙)^(2)=C^(2)-(1-(r_(**))/(r)):}\begin{equation*} \dot{r}^{2}=C^{2}-\left(1-\frac{r_{*}}{r}\right) \tag{3.1611} \end{equation*}(3.1611)r˙2=C2(1rr)
Separating this equation, we find that
(3.1612) d τ = d r C 2 ( 1 r r ) , (3.1612) d τ = d r C 2 1 r r , {:(3.1612)d tau=-(dr)/(sqrt(C^(2)-(1-(r_(**))/(r))))",":}\begin{equation*} d \tau=-\frac{d r}{\sqrt{C^{2}-\left(1-\frac{r_{*}}{r}\right)}}, \tag{3.1612} \end{equation*}(3.1612)dτ=drC2(1rr),
where we fix the sign of the square root so that r ˙ < 0 r ˙ < 0 r^(˙) < 0\dot{r}<0r˙<0, since the particle is falling inward. Thus, we obtain the proper time
(3.1613) Δ τ = r 0 r 1 d r C 2 ( 1 r t r ) (3.1613) Δ τ = r 0 r 1 d r C 2 1 r t r {:(3.1613)Delta tau=-int_(r_(0))^(r_(1))(dr)/(sqrt(C^(2)-(1-(r_(t))/(r)))):}\begin{equation*} \Delta \tau=-\int_{r_{0}}^{r_{1}} \frac{d r}{\sqrt{C^{2}-\left(1-\frac{r_{t}}{r}\right)}} \tag{3.1613} \end{equation*}(3.1613)Δτ=r0r1drC2(1rtr)
where r 0 r 0 r_(0)r_{0}r0 and r 1 r 1 r_(1)r_{1}r1 are the initial and final positions of the particle, respectively, and the integration constant C C CCC is determined by the initial velocity of the particle.
Furthermore, to compute the coordinate time, we have (again fixing r ˙ < 0 r ˙ < 0 r^(˙) < 0\dot{r}<0r˙<0 )
(3.1614) d r d t = r ˙ t ¨ = C 2 ( 1 r r ) 1 C ( 1 r r ) = ( 1 r r ) 1 1 C 2 ( 1 r r ) (3.1614) d r d t = r ˙ t ¨ = C 2 1 r r 1 C 1 r r = 1 r r 1 1 C 2 1 r r {:(3.1614)(dr)/(dt)=((r^(˙)))/((t^(¨)))=-sqrt(C^(2)-(1-(r_(**))/(r)))*(1)/(C)(1-(r_(**))/(r))=-(1-(r_(**))/(r))sqrt(1-(1)/(C^(2))(1-(r_(**))/(r))):}\begin{equation*} \frac{d r}{d t}=\frac{\dot{r}}{\ddot{t}}=-\sqrt{C^{2}-\left(1-\frac{r_{*}}{r}\right)} \cdot \frac{1}{C}\left(1-\frac{r_{*}}{r}\right)=-\left(1-\frac{r_{*}}{r}\right) \sqrt{1-\frac{1}{C^{2}}\left(1-\frac{r_{*}}{r}\right)} \tag{3.1614} \end{equation*}(3.1614)drdt=r˙t¨=C2(1rr)1C(1rr)=(1rr)11C2(1rr)
Thus, integrating, we obtain the coordinate time
(3.1615) Δ t = r 0 r 1 d r ( 1 r r ) 1 1 C 2 ( 1 r r ) (3.1615) Δ t = r 0 r 1 d r 1 r r 1 1 C 2 1 r r {:(3.1615)Delta t=-int_(r_(0))^(r_(1))(dr)/((1-(r_(**))/(r))sqrt(1-(1)/(C^(2))(1-(r_(**))/(r)))):}\begin{equation*} \Delta t=-\int_{r_{0}}^{r_{1}} \frac{d r}{\left(1-\frac{r_{*}}{r}\right) \sqrt{1-\frac{1}{C^{2}}\left(1-\frac{r_{*}}{r}\right)}} \tag{3.1615} \end{equation*}(3.1615)Δt=r0r1dr(1rr)11C2(1rr)
Indeed, the integrand of the integral giving Δ τ Δ τ Delta tau\Delta \tauΔτ remains finite when r r r r r rarrr_(**)r \rightarrow r_{*}rr, actually the limit is 1 / C 1 / C -1//C-1 / C1/C, and thus, Δ τ Δ τ Delta tau\Delta \tauΔτ is finite as r 1 r r 1 r r_(1)rarrr_(**)r_{1} \rightarrow r_{*}r1r, which means that it only takes a finite proper time to reach the even horizon at r r r_(**)r_{*}r. However, the integrand of the integral giving Δ t Δ t Delta t\Delta tΔt diverges as ( r r 1 ) 1 r r 1 1 -((r)/(r_(**))-1)^(-1)-\left(\frac{r}{r_{*}}-1\right)^{-1}(rr1)1 when r r r r r rarrr_(**)r \rightarrow r_{*}rr, and thus, Δ t Δ t Delta t\Delta tΔt diverges as
(3.1616) r 0 r 1 d r r r 1 = r ln r 0 / r 1 r 1 / r 1 (3.1616) r 0 r 1 d r r r 1 = r ln r 0 / r 1 r 1 / r 1 {:(3.1616)-int_(r_(0))^(r_(1))(dr)/((r)/(r_(**))-1)=r_(**)ln((r_(0)//r_(**)-1)/(r_(1)//r_(**)-1)):}\begin{equation*} -\int_{r_{0}}^{r_{1}} \frac{d r}{\frac{r}{r_{*}}-1}=r_{*} \ln \frac{r_{0} / r_{*}-1}{r_{1} / r_{*}-1} \tag{3.1616} \end{equation*}(3.1616)r0r1drrr1=rlnr0/r1r1/r1
when r 1 r r 1 r r_(1)rarrr_(**)r_{1} \rightarrow r_{*}r1r, which means that it takes an infinite amount of coordinate time to reach the event horizon at r r r_(**)r_{*}r.
In order to reach the same conclusions as above, we can choose C = 1 C = 1 C=1C=1C=1 and find explicitly that
Δ τ = r 0 r 1 d r 1 ( 1 r r ) = r 0 r 1 r r d r = 2 3 r ( r 0 3 / 2 r 1 3 / 2 ) (3.1617) 2 3 ( r 0 r 0 r r ) < when r 1 r , Δ t = r 0 r 1 d r ( 1 r r ) 1 ( 1 r r ) = r 0 r 1 1 1 r r r r d r = 2 3 [ ( r 0 + 3 r ) r 0 r ( r 1 + 3 r ) r 1 r 3 r ( artanh r 0 r artanh r 1 r ) ] (3.1618) when r 1 r , Δ τ = r 0 r 1 d r 1 1 r r = r 0 r 1 r r d r = 2 3 r r 0 3 / 2 r 1 3 / 2 (3.1617) 2 3 r 0 r 0 r r <  when  r 1 r , Δ t = r 0 r 1 d r 1 r r 1 1 r r = r 0 r 1 1 1 r r r r d r = 2 3 r 0 + 3 r r 0 r r 1 + 3 r r 1 r 3 r artanh r 0 r artanh r 1 r (3.1618)  when  r 1 r , {:[Delta tau=-int_(r_(0))^(r_(1))(dr)/(sqrt(1-(1-(r_(**))/(r):})))=-int_(r_(0))^(r_(1))sqrt((r)/(r_(**)))dr=(2)/(3sqrt(r_(**)))(r_(0)^(3//2)-r_(1)^(3//2))],[(3.1617) rarr(2)/(3)(r_(0)sqrt({:(r_(0))/(r_(**))-r_(**)) < ooquad" when "r_(1)rarrr_(**),):}],[Delta t=-int_(r_(0))^(r_(1))(dr)/((1-(r_(**))/(r))sqrt(1-(1-(r_(**))/(r))))=-int_(r_(0))^(r_(1))(1)/(1-(r_(**))/(r))sqrt((r)/(r_(**)))dr],[=(2)/(3)[(r_(0)+3r_(**))sqrt((r_(0))/(r_(**)))-(r_(1)+3r_(**))sqrt((r_(1))/(r_(**)))-3r_(**)(artanhsqrt((r_(0))/(r_(**)))-artanhsqrt((r_(1))/(r_(**))))]],[(3.1618) rarr oo" when "r_(1)rarrr_(**)","]:}\begin{align*} \Delta \tau & =-\int_{r_{0}}^{r_{1}} \frac{d r}{\left.\sqrt{1-\left(1-\frac{r_{*}}{r}\right.}\right)}=-\int_{r_{0}}^{r_{1}} \sqrt{\frac{r}{r_{*}}} d r=\frac{2}{3 \sqrt{r_{*}}}\left(r_{0}^{3 / 2}-r_{1}^{3 / 2}\right) \\ & \rightarrow \frac{2}{3}\left(r_{0} \sqrt{\left.\frac{r_{0}}{r_{*}}-r_{*}\right)<\infty \quad \text { when } r_{1} \rightarrow r_{*},}\right. \tag{3.1617}\\ \Delta t & =-\int_{r_{0}}^{r_{1}} \frac{d r}{\left(1-\frac{r_{*}}{r}\right) \sqrt{1-\left(1-\frac{r_{*}}{r}\right)}}=-\int_{r_{0}}^{r_{1}} \frac{1}{1-\frac{r_{*}}{r}} \sqrt{\frac{r}{r_{*}}} d r \\ & =\frac{2}{3}\left[\left(r_{0}+3 r_{*}\right) \sqrt{\frac{r_{0}}{r_{*}}}-\left(r_{1}+3 r_{*}\right) \sqrt{\frac{r_{1}}{r_{*}}}-3 r_{*}\left(\operatorname{artanh} \sqrt{\frac{r_{0}}{r_{*}}}-\operatorname{artanh} \sqrt{\frac{r_{1}}{r_{*}}}\right)\right] \\ & \rightarrow \infty \text { when } r_{1} \rightarrow r_{*}, \tag{3.1618} \end{align*}Δτ=r0r1dr1(1rr)=r0r1rrdr=23r(r03/2r13/2)(3.1617)23(r0r0rr)< when r1r,Δt=r0r1dr(1rr)1(1rr)=r0r111rrrrdr=23[(r0+3r)r0r(r1+3r)r1r3r(artanhr0rartanhr1r)](3.1618) when r1r,
since lim x 1 artanh x lim x 1 artanh x lim_(x rarr1)artanh x rarr oo\lim _{x \rightarrow 1} \operatorname{artanh} x \rightarrow \inftylimx1artanhx. The choice of C = 1 C = 1 C=1C=1C=1 corresponds to t ˙ = ( 1 r r ) 1 t ˙ = 1 r r 1 t^(˙)=(1-(r_(**))/(r))^(-1)rarr\dot{t}=\left(1-\frac{r_{*}}{r}\right)^{-1} \rightarrowt˙=(1rr)1 1 as r r r rarr oor \rightarrow \inftyr, i.e., the kinetic energy of the particle is chosen exactly such that it would be at rest at r r r rarr oor \rightarrow \inftyr.

2.96

Initially, since the observer is moving tangentially, r ˙ = 0 r ˙ = 0 r^(˙)=0\dot{r}=0r˙=0. Let us consider the constants of motion for the observer due to t t del_(t)\partial_{t}t and φ φ del_(varphi)\partial_{\varphi}φ being Killing vector fields and the observer following a timelike geodesic. Thus, we define
(3.1619) E = 1 2 g ( t , γ ˙ ) 2 , L = g ( φ , γ ˙ ) (3.1619) E = 1 2 g t , γ ˙ 2 , L = g φ , γ ˙ {:(3.1619)E=(1)/(2)g(del_(t),(gamma^(˙)))^(2)","quad L=-g(del_(varphi),(gamma^(˙))):}\begin{equation*} E=\frac{1}{2} g\left(\partial_{t}, \dot{\gamma}\right)^{2}, \quad L=-g\left(\partial_{\varphi}, \dot{\gamma}\right) \tag{3.1619} \end{equation*}(3.1619)E=12g(t,γ˙)2,L=g(φ,γ˙)
where γ ˙ γ ˙ gamma^(˙)\dot{\gamma}γ˙ is the observer's 4 -velocity. We use a coordinate system such that θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2 and the initial tangent at r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0 is therefore
(3.1620) γ ˙ 0 = α t + β φ (3.1620) γ ˙ 0 = α t + β φ {:(3.1620)gamma^(˙)_(0)=alphadel_(t)+betadel_(varphi):}\begin{equation*} \dot{\gamma}_{0}=\alpha \partial_{t}+\beta \partial_{\varphi} \tag{3.1620} \end{equation*}(3.1620)γ˙0=αt+βφ
The numbers α α alpha\alphaα and β β beta\betaβ are determined from the normalization of the 4 -velocity and that the velocity relative to the stationary frame is v 0 v 0 v_(0)v_{0}v0 given as
(3.1621) g ( γ ˙ 0 , γ ˙ 0 ) = α 2 ( 1 r r 0 ) β 2 r 0 2 = 1 , 1 1 v 0 2 = g ( V , γ ˙ ) = α A ( 1 r r 0 ) (3.1621) g γ ˙ 0 , γ ˙ 0 = α 2 1 r r 0 β 2 r 0 2 = 1 , 1 1 v 0 2 = g ( V , γ ˙ ) = α A 1 r r 0 {:(3.1621)g(gamma^(˙)_(0),gamma^(˙)_(0))=alpha^(2)(1-(r_(**))/(r_(0)))-beta^(2)r_(0)^(2)=1","quad(1)/(sqrt(1-v_(0)^(2)))=g(V","gamma^(˙))=alpha A(1-(r_(**))/(r_(0))):}\begin{equation*} g\left(\dot{\gamma}_{0}, \dot{\gamma}_{0}\right)=\alpha^{2}\left(1-\frac{r_{*}}{r_{0}}\right)-\beta^{2} r_{0}^{2}=1, \quad \frac{1}{\sqrt{1-v_{0}^{2}}}=g(V, \dot{\gamma})=\alpha A\left(1-\frac{r_{*}}{r_{0}}\right) \tag{3.1621} \end{equation*}(3.1621)g(γ˙0,γ˙0)=α2(1rr0)β2r02=1,11v02=g(V,γ˙)=αA(1rr0)
where V = A t V = A t V=Adel_(t)V=A \partial_{t}V=At is the 4-velocity of a local stationary observer at r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0. From the normalization of V V VVV follows that
(3.1622) g ( V , V ) = A 2 ( 1 r r 0 ) = 1 A = r 0 r 0 r (3.1622) g ( V , V ) = A 2 1 r r 0 = 1 A = r 0 r 0 r {:(3.1622)g(V","V)=A^(2)(1-(r_(**))/(r_(0)))=1quad Longrightarrowquad A=sqrt((r_(0))/(r_(0)-r_(**))):}\begin{equation*} g(V, V)=A^{2}\left(1-\frac{r_{*}}{r_{0}}\right)=1 \quad \Longrightarrow \quad A=\sqrt{\frac{r_{0}}{r_{0}-r_{*}}} \tag{3.1622} \end{equation*}(3.1622)g(V,V)=A2(1rr0)=1A=r0r0r
Solving for α α alpha\alphaα and β β beta\betaβ leads to
(3.1623) α = r 0 ( 1 v 0 2 ) ( r 0 r ) , β = v 0 r 0 1 v 0 2 (3.1623) α = r 0 1 v 0 2 r 0 r , β = v 0 r 0 1 v 0 2 {:(3.1623)alpha=sqrt((r_(0))/((1-v_(0)^(2))(r_(0)-r_(**))))","quad beta=(v_(0))/(r_(0)sqrt(1-v_(0)^(2))):}\begin{equation*} \alpha=\sqrt{\frac{r_{0}}{\left(1-v_{0}^{2}\right)\left(r_{0}-r_{*}\right)}}, \quad \beta=\frac{v_{0}}{r_{0} \sqrt{1-v_{0}^{2}}} \tag{3.1623} \end{equation*}(3.1623)α=r0(1v02)(r0r),β=v0r01v02
The constant of motion L L LLL is now expressed as
(3.1624) L = g ( φ , γ ˙ ) = r 0 2 β = v 0 r 0 1 v 0 2 (3.1624) L = g φ , γ ˙ = r 0 2 β = v 0 r 0 1 v 0 2 {:(3.1624)L=-g(del_(varphi),(gamma^(˙)))=r_(0)^(2)beta=(v_(0)r_(0))/(sqrt(1-v_(0)^(2))):}\begin{equation*} L=-g\left(\partial_{\varphi}, \dot{\gamma}\right)=r_{0}^{2} \beta=\frac{v_{0} r_{0}}{\sqrt{1-v_{0}^{2}}} \tag{3.1624} \end{equation*}(3.1624)L=g(φ,γ˙)=r02β=v0r01v02
Examining the effective potential
(3.1625) V ( r ) = 1 2 ( 1 + L 2 r 2 ) ( 1 r r ) (3.1625) V ( r ) = 1 2 1 + L 2 r 2 1 r r {:(3.1625)V(r)=(1)/(2)(1+(L^(2))/(r^(2)))(1-(r_(**))/(r)):}\begin{equation*} V(r)=\frac{1}{2}\left(1+\frac{L^{2}}{r^{2}}\right)\left(1-\frac{r_{*}}{r}\right) \tag{3.1625} \end{equation*}(3.1625)V(r)=12(1+L2r2)(1rr)
there are two possible situations when the observer will not fall into the black hole
  1. When d V / d r 0 d V / d r 0 dV//dr <= 0d V / d r \leq 0dV/dr0, the observer will initially start moving outward. The position r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0 will then always be a turning point meaning that r r 0 r r 0 r >= r_(0)r \geq r_{0}rr0 for the entire solution. In the case of d V / d r = 0 d V / d r = 0 dV//dr=0d V / d r=0dV/dr=0, the observer will move in a circular orbit.
  2. When d V / d r > 0 d V / d r > 0 dV//dr > 0d V / d r>0dV/dr>0, the observer will start moving toward the black hole. The requirement not to fall into the black hole is then that there is a turning point at a smaller value of r r rrr, i.e., that V ( r ) = V ( r 0 ) V ( r ) = V r 0 V(r)=V(r_(0))V(r)=V\left(r_{0}\right)V(r)=V(r0) for some r < r 0 r < r 0 r < r_(0)r<r_{0}r<r0. There will then necessarily be another turning point, which is not accessible, at an even smaller r r rrr. The border of this possibility occurs when both of the turning points coincide, i.e., when V ( x ) V ( x 0 ) V ( x ) V x 0 V(x)-V(x_(0))V(x)-V\left(x_{0}\right)V(x)V(x0) has a double root different from x 0 x 0 x_(0)x_{0}x0.
To see when the first case to applies, we compute d V / d x d V / d x dV//dxd V / d xdV/dx with x = 1 / r x = 1 / r x=1//rx=1 / rx=1/r. As d x / d r < 0 , d V / d r 0 d x / d r < 0 , d V / d r 0 dx//dr < 0,dV//dr <= 0d x / d r<0, d V / d r \leq 0dx/dr<0,dV/dr0 is equivalent with d V / d x = ( d V / d r ) ( d r / d x ) 0 d V / d x = ( d V / d r ) ( d r / d x ) 0 dV//dx=(dV//dr)(dr//dx) >= 0d V / d x=(d V / d r)(d r / d x) \geq 0dV/dx=(dV/dr)(dr/dx)0. We find that
(3.1626) d V d x | x = x 0 = v 0 2 r 0 1 v 0 2 ( 1 r r 0 ) r 2 ( 1 v 0 2 ) 0 v 0 2 r 2 ( r 0 r ) (3.1626) d V d x x = x 0 = v 0 2 r 0 1 v 0 2 1 r r 0 r 2 1 v 0 2 0 v 0 2 r 2 r 0 r {:(3.1626)(dV)/(dx)|_(x=x_(0))=(v_(0)^(2)r_(0))/(1-v_(0)^(2))(1-(r_(**))/(r_(0)))-(r_(**))/(2(1-v_(0)^(2))) >= 0quad Longrightarrowquadv_(0)^(2) >= (r_(**))/(2(r_(0)-r_(**))):}\begin{equation*} \left.\frac{d V}{d x}\right|_{x=x_{0}}=\frac{v_{0}^{2} r_{0}}{1-v_{0}^{2}}\left(1-\frac{r_{*}}{r_{0}}\right)-\frac{r_{*}}{2\left(1-v_{0}^{2}\right)} \geq 0 \quad \Longrightarrow \quad v_{0}^{2} \geq \frac{r_{*}}{2\left(r_{0}-r_{*}\right)} \tag{3.1626} \end{equation*}(3.1626)dVdx|x=x0=v02r01v02(1rr0)r2(1v02)0v02r2(r0r)
This puts a lower limit on the velocity v 0 v 0 v_(0)v_{0}v0 for which case 1 applies.
Case 2 is a bit more complicated. In order for three turning points to exist, we must have three real solutions to V ( x ) V ( x 0 ) = 0 V ( x ) V x 0 = 0 V(x)-V(x_(0))=0V(x)-V\left(x_{0}\right)=0V(x)V(x0)=0, where the left-hand side is a third degree polynomial. Luckily, we know that x = x 0 x = x 0 x=x_(0)x=x_{0}x=x0 is a root of the polynomial and can therefore be factored out and we are left with having to have two real roots for a second degree polynomial. After some algebra, we find that the requirement on v 0 v 0 v_(0)v_{0}v0 for all roots to be real is
(3.1627) v 0 2 r 2 r 0 + R (3.1627) v 0 2 r 2 r 0 + R {:(3.1627)v_(0) >= (2r_(**)^(2))/(r_(0)+R):}\begin{equation*} v_{0} \geq \frac{2 r_{*}^{2}}{r_{0}+R} \tag{3.1627} \end{equation*}(3.1627)v02r2r0+R
However, we also need to ensure that r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0 is the turning point at the largest r r rrr. This is the case whenever d V / d x < 0 d V / d x < 0 dV//dx < 0d V / d x<0dV/dx<0 and the midpoint of the other solutions is smaller than r 0 r 0 r_(0)r_{0}r0. The midpoint 1 / r ¯ 1 / r ¯ 1// bar(r)1 / \bar{r}1/r¯ is found to be located at
(3.1628) 1 r ¯ = 1 2 ( 1 r 1 r 0 ) (3.1628) 1 r ¯ = 1 2 1 r 1 r 0 {:(3.1628)(1)/(( bar(r)))=(1)/(2)((1)/(r_(**))-(1)/(r_(0))):}\begin{equation*} \frac{1}{\bar{r}}=\frac{1}{2}\left(\frac{1}{r_{*}}-\frac{1}{r_{0}}\right) \tag{3.1628} \end{equation*}(3.1628)1r¯=12(1r1r0)
Requiring that r 0 > r ¯ r 0 > r ¯ r_(0) > bar(r)r_{0}>\bar{r}r0>r¯ then leads to
(3.1629) r 0 > 3 r (3.1629) r 0 > 3 r {:(3.1629)r_(0) > 3r_(**):}\begin{equation*} r_{0}>3 r_{*} \tag{3.1629} \end{equation*}(3.1629)r0>3r
Thus, in the region r 0 3 r r 0 3 r r_(0) <= 3r_(**)r_{0} \leq 3 r_{*}r03r, case 1 always applies and for r 0 > 3 r r 0 > 3 r r_(0) > 3r_(**)r_{0}>3 r_{*}r0>3r, we find that the requirement not to fall into the black hole is given by
(3.1630) v 0 2 min ( 4 r 2 ( r 0 + r ) 2 , r 2 ( r 0 r ) ) = 4 r 2 ( r 0 + r ) 2 (3.1630) v 0 2 min 4 r 2 r 0 + r 2 , r 2 r 0 r = 4 r 2 r 0 + r 2 {:(3.1630)v_(0)^(2) >= min((4r_(**)^(2))/((r_(0)+r_(**))^(2)),(r_(**))/(2(r_(0)-r_(**))))=(4r_(**)^(2))/((r_(0)+r_(**))^(2)):}\begin{equation*} v_{0}^{2} \geq \min \left(\frac{4 r_{*}^{2}}{\left(r_{0}+r_{*}\right)^{2}}, \frac{r_{*}}{2\left(r_{0}-r_{*}\right)}\right)=\frac{4 r_{*}^{2}}{\left(r_{0}+r_{*}\right)^{2}} \tag{3.1630} \end{equation*}(3.1630)v02min(4r2(r0+r)2,r2(r0r))=4r2(r0+r)2
(The expressions are equal when r 0 = 3 r r 0 = 3 r r_(0)=3r_(**)r_{0}=3 r_{*}r0=3r and otherwise the first expression is always smaller.) The solution to the given problem is therefore that the minimal velocity v 0 v 0 v_(0)v_{0}v0 is given by
(3.1631) v 0 , min = { r 2 ( r 0 r ) , r 0 3 r 2 r r 0 + r , r 0 3 r (3.1631) v 0 , min = r 2 r 0 r , r 0 3 r 2 r r 0 + r , r 0 3 r {:(3.1631)v_(0,min)={[sqrt((r_(**))/(2(r_(0)-r_(**))))",",r_(0) <= 3r_(**)],[(2r_(**))/(r_(0)+r_(**))",",r_(0) >= 3r_(**)]:}:}v_{0, \min }= \begin{cases}\sqrt{\frac{r_{*}}{2\left(r_{0}-r_{*}\right)}}, & r_{0} \leq 3 r_{*} \tag{3.1631}\\ \frac{2 r_{*}}{r_{0}+r_{*}}, & r_{0} \geq 3 r_{*}\end{cases}(3.1631)v0,min={r2(r0r),r03r2rr0+r,r03r
b) For v 0 = 0 v 0 = 0 v_(0)=0v_{0}=0v0=0, we have the effective potential
(3.1632) V ( r ) = 1 2 ( 1 r r ) (3.1632) V ( r ) = 1 2 1 r r {:(3.1632)V(r)=(1)/(2)(1-(r_(**))/(r)):}\begin{equation*} V(r)=\frac{1}{2}\left(1-\frac{r_{*}}{r}\right) \tag{3.1632} \end{equation*}(3.1632)V(r)=12(1rr)
since L = 0 L = 0 L=0L=0L=0. In addition, the initial 4 -velocity is equal to that of the stationary observer, which was found in a), i.e.,
(3.1633) γ ˙ 0 = r 0 r 0 r t (3.1633) γ ˙ 0 = r 0 r 0 r t {:(3.1633)gamma^(˙)_(0)=sqrt((r_(0))/(r_(0)-r_(**)))del_(t):}\begin{equation*} \dot{\gamma}_{0}=\sqrt{\frac{r_{0}}{r_{0}-r_{*}}} \partial_{t} \tag{3.1633} \end{equation*}(3.1633)γ˙0=r0r0rt
Evaluated at r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0, the constant of motion is therefore
(3.1634) E = 1 2 ( 1 r r 0 ) (3.1634) E = 1 2 1 r r 0 {:(3.1634)E=(1)/(2)(1-(r_(**))/(r_(0))):}\begin{equation*} E=\frac{1}{2}\left(1-\frac{r_{*}}{r_{0}}\right) \tag{3.1634} \end{equation*}(3.1634)E=12(1rr0)
The equation of motion for the r r rrr-coordinate is then
(3.1635) r ˙ 2 = 2 E 2 V ( r ) = r r r r 0 d r r r r r 0 = d τ (3.1635) r ˙ 2 = 2 E 2 V ( r ) = r r r r 0 d r r r r r 0 = d τ {:(3.1635)r^(˙)^(2)=2E-2V(r)=(r_(**))/(r)-(r_(**))/(r_(0))Longrightarrow(-dr)/(sqrt((r_(**))/(r)-(r_(**))/(r_(0))))=d tau:}\begin{equation*} \dot{r}^{2}=2 E-2 V(r)=\frac{r_{*}}{r}-\frac{r_{*}}{r_{0}} \Longrightarrow \frac{-d r}{\sqrt{\frac{r_{*}}{r}-\frac{r_{*}}{r_{0}}}}=d \tau \tag{3.1635} \end{equation*}(3.1635)r˙2=2E2V(r)=rrrr0drrrrr0=dτ
where we have used that d r d r drd rdr is negative to omit the positive root. The proper time to reach the singularity is given by integrating this from r 0 r 0 r_(0)r_{0}r0 to 0 and we find
(3.1636) τ = r 0 0 d r r r r r 0 = 1 r x 0 d x x 2 x x 0 = π r 0 2 r 0 r (3.1636) τ = r 0 0 d r r r r r 0 = 1 r x 0 d x x 2 x x 0 = π r 0 2 r 0 r {:(3.1636)tau=-int_(r_(0))^(0)(dr)/(sqrt((r_(**))/(r)-(r_(**))/(r_(0))))=(1)/(sqrt(r_(**)))int_(x_(0))^(oo)(dx)/(x^(2)sqrt(x-x_(0)))=(pir_(0))/(2)sqrt((r_(0))/(r_(**))):}\begin{equation*} \tau=-\int_{r_{0}}^{0} \frac{d r}{\sqrt{\frac{r_{*}}{r}-\frac{r_{*}}{r_{0}}}}=\frac{1}{\sqrt{r_{*}}} \int_{x_{0}}^{\infty} \frac{d x}{x^{2} \sqrt{x-x_{0}}}=\frac{\pi r_{0}}{2} \sqrt{\frac{r_{0}}{r_{*}}} \tag{3.1636} \end{equation*}(3.1636)τ=r00drrrrr0=1rx0dxx2xx0=πr02r0r
Note that the integral is applicable even across the Schwarzschild event horizon as its form removes the coordinate singularity that appears there. The integral itself therefore does not suffer from the coordinate singularity and behaves nicely all the way to the singularity at r = 0 r = 0 r=0r=0r=0.

2.97

When the engines are turned off, the satellite will follow a geodesic. Assuming θ = π / 2 = θ = π / 2 = theta=pi//2=\theta=\pi / 2=θ=π/2= const., the Lagrangian is given by
(3.1637) L = g μ ν x ˙ μ x ˙ ν = ( 1 r r ) t ˙ 2 ( 1 r r ) 1 r ˙ 2 r 2 ϕ ˙ 2 (3.1637) L = g μ ν x ˙ μ x ˙ ν = 1 r r t ˙ 2 1 r r 1 r ˙ 2 r 2 ϕ ˙ 2 {:(3.1637)L=g_(mu nu)x^(˙)^(mu)x^(˙)^(nu)=(1-(r_(**))/(r))t^(˙)^(2)-(1-(r_(**))/(r))^(-1)r^(˙)^(2)-r^(2)phi^(˙)^(2):}\begin{equation*} \mathcal{L}=g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=\left(1-\frac{r_{*}}{r}\right) \dot{t}^{2}-\left(1-\frac{r_{*}}{r}\right)^{-1} \dot{r}^{2}-r^{2} \dot{\phi}^{2} \tag{3.1637} \end{equation*}(3.1637)L=gμνx˙μx˙ν=(1rr)t˙2(1rr)1r˙2r2ϕ˙2
No t t ttt and ϕ ϕ phi\phiϕ dependencies give two constants of motions, coming from the EulerLagrange equations, i.e.,
(3.1638) κ = ( 1 r r ) t ˙ , = r 2 ϕ ˙ = R 2 B (3.1638) κ = 1 r r t ˙ , = r 2 ϕ ˙ = R 2 B {:(3.1638)kappa=(1-(r_(**))/(r))t^(˙)","quadℓ=r^(2)phi^(˙)=R^(2)B:}\begin{equation*} \kappa=\left(1-\frac{r_{*}}{r}\right) \dot{t}, \quad \ell=r^{2} \dot{\phi}=R^{2} B \tag{3.1638} \end{equation*}(3.1638)κ=(1rr)t˙,=r2ϕ˙=R2B
We also get a third relation from setting L = 1 L = 1 L=1\mathcal{L}=1L=1, namely
(3.1639) 1 = ( 1 r r ) t ˙ 2 ( 1 r r ) 1 r ˙ 2 r 2 ϕ ˙ 2 (3.1639) 1 = 1 r r t ˙ 2 1 r r 1 r ˙ 2 r 2 ϕ ˙ 2 {:(3.1639)1=(1-(r_(**))/(r))t^(˙)^(2)-(1-(r_(**))/(r))^(-1)r^(˙)^(2)-r^(2)phi^(˙)^(2):}\begin{equation*} 1=\left(1-\frac{r_{*}}{r}\right) \dot{t}^{2}-\left(1-\frac{r_{*}}{r}\right)^{-1} \dot{r}^{2}-r^{2} \dot{\phi}^{2} \tag{3.1639} \end{equation*}(3.1639)1=(1rr)t˙2(1rr)1r˙2r2ϕ˙2
Using the first and second equations, we can rewrite the third equation in terms of just radial dependence (with r ˙ = 0 r ˙ = 0 r^(˙)=0\dot{r}=0r˙=0 ):
1 = ( 1 r r ) 1 κ 2 r 2 2 = ( 1 r r ) 1 κ 2 R 4 B 2 r 2 (3.1640) κ 2 = ( 1 r r ) ( 1 + R 4 B 2 r 2 ) 1 = 1 r r 1 κ 2 r 2 2 = 1 r r 1 κ 2 R 4 B 2 r 2 (3.1640) κ 2 = 1 r r 1 + R 4 B 2 r 2 {:[1=(1-(r_(**))/(r))^(-1)kappa^(2)-r^(-2)ℓ^(2)=(1-(r_(**))/(r))^(-1)kappa^(2)-(R^(4)B^(2))/(r^(2))],[(3.1640)=>kappa^(2)=(1-(r_(**))/(r))(1+(R^(4)B^(2))/(r^(2)))]:}\begin{align*} 1 & =\left(1-\frac{r_{*}}{r}\right)^{-1} \kappa^{2}-r^{-2} \ell^{2}=\left(1-\frac{r_{*}}{r}\right)^{-1} \kappa^{2}-\frac{R^{4} B^{2}}{r^{2}} \\ & \Rightarrow \kappa^{2}=\left(1-\frac{r_{*}}{r}\right)\left(1+\frac{R^{4} B^{2}}{r^{2}}\right) \tag{3.1640} \end{align*}1=(1rr)1κ2r22=(1rr)1κ2R4B2r2(3.1640)κ2=(1rr)(1+R4B2r2)
Normally, these three equations are sufficient, and the third equation is equivalent to the equation of motion that one obtains from varying the Lagrangian with respect to r r rrr, but it happens that there is an exception to this case and that is precisely when
r ˙ = 0 r ˙ = 0 r^(˙)=0\dot{r}=0r˙=0. Varying the Lagrangian with respect to r r rrr and neglecting all time derivatives of r r rrr gives the equation
(3.1641) r r 2 ( 1 r r ) 2 κ 2 2 r ( R 2 B r 2 ) 2 = 0 κ 2 = 2 r 3 r ( 1 r r ) 2 ( R 2 B r 2 ) 2 (3.1641) r r 2 1 r r 2 κ 2 2 r R 2 B r 2 2 = 0 κ 2 = 2 r 3 r 1 r r 2 R 2 B r 2 2 {:(3.1641)(r_(**))/(r^(2))(1-(r_(**))/(r))^(-2)kappa^(2)-2r((R^(2)B)/(r^(2)))^(2)=0=>kappa^(2)=(2r^(3))/(r_(**))(1-(r_(**))/(r))^(2)((R^(2)B)/(r^(2)))^(2):}\begin{equation*} \frac{r_{*}}{r^{2}}\left(1-\frac{r_{*}}{r}\right)^{-2} \kappa^{2}-2 r\left(\frac{R^{2} B}{r^{2}}\right)^{2}=0 \Rightarrow \kappa^{2}=\frac{2 r^{3}}{r_{*}}\left(1-\frac{r_{*}}{r}\right)^{2}\left(\frac{R^{2} B}{r^{2}}\right)^{2} \tag{3.1641} \end{equation*}(3.1641)rr2(1rr)2κ22r(R2Br2)2=0κ2=2r3r(1rr)2(R2Br2)2
Now, at r = R r = R r=Rr=Rr=R, we find B B BBB according to
(3.1642) ( 1 r R ) ( 1 + R 2 B 2 ) = 2 R 3 B 2 r ( 1 r R ) 2 B 2 = R 2 ( 2 R r 3 ) 1 (3.1642) 1 r R 1 + R 2 B 2 = 2 R 3 B 2 r 1 r R 2 B 2 = R 2 2 R r 3 1 {:(3.1642)(1-(r_(**))/(R))(1+R^(2)B^(2))=(2R^(3)B^(2))/(r_(**))(1-(r_(**))/(R))^(2)quad=>quadB^(2)=R^(-2)((2R)/(r_(**))-3)^(-1):}\begin{equation*} \left(1-\frac{r_{*}}{R}\right)\left(1+R^{2} B^{2}\right)=\frac{2 R^{3} B^{2}}{r_{*}}\left(1-\frac{r_{*}}{R}\right)^{2} \quad \Rightarrow \quad B^{2}=R^{-2}\left(\frac{2 R}{r_{*}}-3\right)^{-1} \tag{3.1642} \end{equation*}(3.1642)(1rR)(1+R2B2)=2R3B2r(1rR)2B2=R2(2Rr3)1
We make an interesting observation. We have a minimal radius for which we can have a geodesic circular orbit around a star or a black hole, i.e., R = 3 r / 2 R = 3 r / 2 R=3r_(**)//2R=3 r_{*} / 2R=3r/2. This orbit will be unstable. Stable circular orbits only exist when the radius R R RRR is larger than 3 r / 2 3 r / 2 3r_(**)//23 r_{*} / 23r/2, i.e., R > 3 r / 2 R > 3 r / 2 R > 3r_(**)//2R>3 r_{*} / 2R>3r/2. For the case of the Earth, R > R 0 3 r / 2 R > R 0 3 r / 2 R > R_(0)≫3r_(**)//2R>R_{0} \gg 3 r_{*} / 2R>R03r/2.

2.98

The satellite's orbit is at constant radial distance r r rrr from the Earth, which means that d r = 0 d r = 0 dr=0d r=0dr=0.
a) Now, the proper time for the satellite is given by
(3.1643) d τ 2 = d s 2 (3.1643) d τ 2 = d s 2 {:(3.1643)dtau^(2)=ds^(2):}\begin{equation*} d \tau^{2}=d s^{2} \tag{3.1643} \end{equation*}(3.1643)dτ2=ds2
Using the Schwarzschild metric with d r = 0 d r = 0 dr=0d r=0dr=0 and d θ = 0 d θ = 0 d theta=0d \theta=0dθ=0 (since θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2 ) for the satellite's orbit around the Earth and that v = r d ϕ / d t v = r d ϕ / d t v=rd phi//dtv=r d \phi / d tv=rdϕ/dt, we find that
d τ 2 = ( 1 r r ) d t 2 r 2 d ϕ 2 = ( 1 r r ) d t 2 ( r d ϕ d t ) 2 d t 2 (3.1644) = [ ( 1 r r ) v 2 ] d t 2 d τ 2 = 1 r r d t 2 r 2 d ϕ 2 = 1 r r d t 2 r d ϕ d t 2 d t 2 (3.1644) = 1 r r v 2 d t 2 {:[dtau^(2)=(1-(r_(**))/(r))dt^(2)-r^(2)dphi^(2)=(1-(r_(**))/(r))dt^(2)-(r(d phi)/(dt))^(2)dt^(2)],[(3.1644)=[(1-(r_(**))/(r))-v^(2)]dt^(2)]:}\begin{align*} d \tau^{2}=\left(1-\frac{r_{*}}{r}\right) d t^{2}-r^{2} d \phi^{2} & =\left(1-\frac{r_{*}}{r}\right) d t^{2}-\left(r \frac{d \phi}{d t}\right)^{2} d t^{2} \\ & =\left[\left(1-\frac{r_{*}}{r}\right)-v^{2}\right] d t^{2} \tag{3.1644} \end{align*}dτ2=(1rr)dt2r2dϕ2=(1rr)dt2(rdϕdt)2dt2(3.1644)=[(1rr)v2]dt2
which implies that
(3.1645) d τ = ± ( 1 r r ) v 2 d t (3.1645) d τ = ± 1 r r v 2 d t {:(3.1645)d tau=+-sqrt((1-(r_(**))/(r))-v^(2))dt:}\begin{equation*} d \tau= \pm \sqrt{\left(1-\frac{r_{*}}{r}\right)-v^{2}} d t \tag{3.1645} \end{equation*}(3.1645)dτ=±(1rr)v2dt
where the square root is independent of time t t ttt. Thus, using d t / d τ > 0 d t / d τ > 0 dt//d tau > 0d t / d \tau>0dt/dτ>0 and integrating from t = 0 t = 0 t=0t=0t=0 to t = T = 2 π r / v t = T = 2 π r / v t=T=2pi r//vt=T=2 \pi r / vt=T=2πr/v, we obtain the proper time τ τ tau\tauτ for the satellite to complete one orbit around the Earth as
(3.1646) τ = 0 T ( 1 r r ) v 2 d t = T ( 1 r r ) v 2 = 2 π r v ( 1 2 G M r ) v 2 (3.1646) τ = 0 T 1 r r v 2 d t = T 1 r r v 2 = 2 π r v 1 2 G M r v 2 {:(3.1646)tau=int_(0)^(T)sqrt((1-(r_(**))/(r))-v^(2))dt=Tsqrt((1-(r_(**))/(r))-v^(2))=(2pi r)/(v)sqrt((1-(2GM)/(r))-v^(2)):}\begin{equation*} \tau=\int_{0}^{T} \sqrt{\left(1-\frac{r_{*}}{r}\right)-v^{2}} d t=T \sqrt{\left(1-\frac{r_{*}}{r}\right)-v^{2}}=\frac{2 \pi r}{v} \sqrt{\left(1-\frac{2 G M}{r}\right)-v^{2}} \tag{3.1646} \end{equation*}(3.1646)τ=0T(1rr)v2dt=T(1rr)v2=2πrv(12GMr)v2
b) The gravitational potential at the satellite is given by
(3.1647) Φ s = 1 2 r r = G M r (3.1647) Φ s = 1 2 r r = G M r {:(3.1647)Phi_(s)=-(1)/(2)(r_(**))/(r)=-(GM)/(r):}\begin{equation*} \Phi_{s}=-\frac{1}{2} \frac{r_{*}}{r}=-\frac{G M}{r} \tag{3.1647} \end{equation*}(3.1647)Φs=12rr=GMr
Now, we calculate the ratio between the coordinate time t = T t = T t=Tt=Tt=T and the proper time for the satellite τ τ tau\tauτ. Then, we series expand this ratio for small v v vvv and r / r r / r r_(**)//rr_{*} / rr/r. Therefore, using the result in a) for T / τ T / τ T//tauT / \tauT/τ, we find that
(3.1648) T τ = 1 ( 1 r r ) v 2 1 1 r 2 r v 2 2 1 Φ s + v 2 2 (3.1648) T τ = 1 1 r r v 2 1 1 r 2 r v 2 2 1 Φ s + v 2 2 {:(3.1648)(T)/( tau)=(1)/(sqrt((1-(r_(**))/(r))-v^(2)))≃(1)/(1-(r_(**))/(2r)-(v^(2))/(2))≃1-Phi_(s)+(v^(2))/(2):}\begin{equation*} \frac{T}{\tau}=\frac{1}{\sqrt{\left(1-\frac{r_{*}}{r}\right)-v^{2}}} \simeq \frac{1}{1-\frac{r_{*}}{2 r}-\frac{v^{2}}{2}} \simeq 1-\Phi_{s}+\frac{v^{2}}{2} \tag{3.1648} \end{equation*}(3.1648)Tτ=1(1rr)v211r2rv221Φs+v22
Thus, we obtain
(3.1649) T τ 1 v 2 2 Φ s (3.1649) T τ 1 v 2 2 Φ s {:(3.1649)(T)/( tau)-1≃(v^(2))/(2)-Phi_(s):}\begin{equation*} \frac{T}{\tau}-1 \simeq \frac{v^{2}}{2}-\Phi_{s} \tag{3.1649} \end{equation*}(3.1649)Tτ1v22Φs
which is what we wanted to show.

2.99

a) The worldline of the satellite will be given by
(3.1650) t = α s , r = r 0 , φ = β s (3.1650) t = α s , r = r 0 , φ = β s {:(3.1650)t=alpha s","quad r=r_(0)","quad varphi=beta s:}\begin{equation*} t=\alpha s, \quad r=r_{0}, \quad \varphi=\beta s \tag{3.1650} \end{equation*}(3.1650)t=αs,r=r0,φ=βs
as the 4 -velocity U = α t + β φ U = α t + β φ U=alphadel_(t)+betadel_(varphi)U=\alpha \partial_{t}+\beta \partial_{\varphi}U=αt+βφ, with α α alpha\alphaα and β β beta\betaβ constant. It follows that the 4-acceleration is
(3.1651) A = U U = ( α 2 Γ t t a + 2 α β Γ t φ a + β 2 Γ φ φ a ) a = 0 (3.1651) A = U U = α 2 Γ t t a + 2 α β Γ t φ a + β 2 Γ φ φ a a = 0 {:(3.1651)A=grad_(U)U=(alpha^(2)Gamma_(tt)^(a)+2alpha betaGamma_(t varphi)^(a)+beta^(2)Gamma_(varphi varphi)^(a))del_(a)=0:}\begin{equation*} A=\nabla_{U} U=\left(\alpha^{2} \Gamma_{t t}^{a}+2 \alpha \beta \Gamma_{t \varphi}^{a}+\beta^{2} \Gamma_{\varphi \varphi}^{a}\right) \partial_{a}=0 \tag{3.1651} \end{equation*}(3.1651)A=UU=(α2Γtta+2αβΓtφa+β2Γφφa)a=0
since the satellite is in free fall. Given the Christoffel symbols of the Schwarzschild metric, the only nontrivial component of this equation is the a = r a = r a=ra=ra=r component
(3.1652) α 2 Γ t t r + 2 α β Γ t φ r + β 2 Γ φ φ r = α 2 r ( r 0 r ) 2 r 0 3 β 2 ( r 0 r ) = 0 (3.1652) α 2 Γ t t r + 2 α β Γ t φ r + β 2 Γ φ φ r = α 2 r r 0 r 2 r 0 3 β 2 r 0 r = 0 {:(3.1652)alpha^(2)Gamma_(tt)^(r)+2alpha betaGamma_(t varphi)^(r)+beta^(2)Gamma_(varphi varphi)^(r)=alpha^(2)(r_(**)(r_(0)-r_(**)))/(2r_(0)^(3))-beta^(2)(r_(0)-r_(**))=0:}\begin{equation*} \alpha^{2} \Gamma_{t t}^{r}+2 \alpha \beta \Gamma_{t \varphi}^{r}+\beta^{2} \Gamma_{\varphi \varphi}^{r}=\alpha^{2} \frac{r_{*}\left(r_{0}-r_{*}\right)}{2 r_{0}^{3}}-\beta^{2}\left(r_{0}-r_{*}\right)=0 \tag{3.1652} \end{equation*}(3.1652)α2Γttr+2αβΓtφr+β2Γφφr=α2r(r0r)2r03β2(r0r)=0
This therefore leads to
(3.1653) β = α r 2 r 0 3 (3.1653) β = α r 2 r 0 3 {:(3.1653)beta=alphasqrt((r_(**))/(2r_(0)^(3))):}\begin{equation*} \beta=\alpha \sqrt{\frac{r_{*}}{2 r_{0}^{3}}} \tag{3.1653} \end{equation*}(3.1653)β=αr2r03
where we have chosen coordinates such that φ φ varphi\varphiφ increases with proper time. Also requiring that the 4 -velocity is normalized, i.e.,
(3.1654) g ( U , U ) = ( 1 r r 0 ) α 2 r 0 2 β 2 = 1 (3.1654) g ( U , U ) = 1 r r 0 α 2 r 0 2 β 2 = 1 {:(3.1654)g(U","U)=(1-(r_(**))/(r_(0)))alpha^(2)-r_(0)^(2)beta^(2)=1:}\begin{equation*} g(U, U)=\left(1-\frac{r_{*}}{r_{0}}\right) \alpha^{2}-r_{0}^{2} \beta^{2}=1 \tag{3.1654} \end{equation*}(3.1654)g(U,U)=(1rr0)α2r02β2=1
now implies that
(3.1655) α = 1 1 3 r 2 r 0 , β = 1 r 0 r 2 r 0 3 r (3.1655) α = 1 1 3 r 2 r 0 , β = 1 r 0 r 2 r 0 3 r {:(3.1655)alpha=(1)/(sqrt(1-(3r_(**))/(2r_(0))))","quad beta=(1)/(r_(0))sqrt((r_(**))/(2r_(0)-3r_(**))):}\begin{equation*} \alpha=\frac{1}{\sqrt{1-\frac{3 r_{*}}{2 r_{0}}}}, \quad \beta=\frac{1}{r_{0}} \sqrt{\frac{r_{*}}{2 r_{0}-3 r_{*}}} \tag{3.1655} \end{equation*}(3.1655)α=113r2r0,β=1r0r2r03r
b) For the satellite to complete a full lap, φ φ varphi\varphiφ needs to change by 2 π 2 π 2pi2 \pi2π. The proper time Δ s Δ s Delta s\Delta sΔs it takes for the satellite to travel an angle Δ φ Δ φ Delta varphi\Delta \varphiΔφ is given by
(3.1656) Δ φ Δ s = φ ˙ Δ s = Δ φ φ ˙ = Δ φ β . (3.1656) Δ φ Δ s = φ ˙ Δ s = Δ φ φ ˙ = Δ φ β . {:(3.1656)(Delta varphi)/(Delta s)=varphi^(˙)quad Longrightarrowquad Delta s=(Delta varphi)/((varphi^(˙)))=(Delta varphi)/(beta).:}\begin{equation*} \frac{\Delta \varphi}{\Delta s}=\dot{\varphi} \quad \Longrightarrow \quad \Delta s=\frac{\Delta \varphi}{\dot{\varphi}}=\frac{\Delta \varphi}{\beta} . \tag{3.1656} \end{equation*}(3.1656)ΔφΔs=φ˙Δs=Δφφ˙=Δφβ.
For Δ φ = 2 π Δ φ = 2 π Delta varphi=2pi\Delta \varphi=2 \piΔφ=2π, we therefore obtain
(3.1657) Δ s = 2 π r 0 2 r 0 3 r r (3.1657) Δ s = 2 π r 0 2 r 0 3 r r {:(3.1657)Delta s=2pir_(0)sqrt((2r_(0)-3r_(**))/(r_(**))):}\begin{equation*} \Delta s=2 \pi r_{0} \sqrt{\frac{2 r_{0}-3 r_{*}}{r_{*}}} \tag{3.1657} \end{equation*}(3.1657)Δs=2πr02r03rr
c) The stationary observer has 4-velocity V = α 0 t V = α 0 t V=alpha_(0)del_(t)V=\alpha_{0} \partial_{t}V=α0t. From g ( V , V ) = 1 g ( V , V ) = 1 g(V,V)=1g(V, V)=1g(V,V)=1 follows that
(3.1658) α 0 = r 0 r 0 r (3.1658) α 0 = r 0 r 0 r {:(3.1658)alpha_(0)=sqrt((r_(0))/(r_(0)-r_(**))):}\begin{equation*} \alpha_{0}=\sqrt{\frac{r_{0}}{r_{0}-r_{*}}} \tag{3.1658} \end{equation*}(3.1658)α0=r0r0r
This implies that the relative gamma factor between the satellite and the observer is given by
(3.1659) γ = g ( U , V ) = ( 1 r r 0 ) α α 0 = 2 r 0 2 r 2 r 0 3 r = 1 1 v 2 (3.1659) γ = g ( U , V ) = 1 r r 0 α α 0 = 2 r 0 2 r 2 r 0 3 r = 1 1 v 2 {:(3.1659)gamma=g(U","V)=(1-(r_(**))/(r_(0)))alphaalpha_(0)=sqrt((2r_(0)-2r_(**))/(2r_(0)-3r_(**)))=(1)/(sqrt(1-v^(2))):}\begin{equation*} \gamma=g(U, V)=\left(1-\frac{r_{*}}{r_{0}}\right) \alpha \alpha_{0}=\sqrt{\frac{2 r_{0}-2 r_{*}}{2 r_{0}-3 r_{*}}}=\frac{1}{\sqrt{1-v^{2}}} \tag{3.1659} \end{equation*}(3.1659)γ=g(U,V)=(1rr0)αα0=2r02r2r03r=11v2
where v v vvv is the relative speed. Solving for v v vvv results in
(3.1660) v = r 2 ( r 0 r ) (3.1660) v = r 2 r 0 r {:(3.1660)v=sqrt((r_(**))/(2(r_(0)-r_(**)))):}\begin{equation*} v=\sqrt{\frac{r_{*}}{2\left(r_{0}-r_{*}\right)}} \tag{3.1660} \end{equation*}(3.1660)v=r2(r0r)

2.100

a) The static observer (1) has constant spatial coordinates. It is therefore convenient to use a parametrization of its worldline in terms of its proper time s 1 s 1 s_(1)s_{1}s1, which will be proportional to the coordinate time t t ttt
(3.1661) t 1 = α s 1 , r 1 = r 0 , φ 1 = φ 0 (3.1661) t 1 = α s 1 , r 1 = r 0 , φ 1 = φ 0 {:(3.1661)t_(1)=alphas_(1)","quadr_(1)=r_(0)","quadvarphi_(1)=varphi_(0):}\begin{equation*} t_{1}=\alpha s_{1}, \quad r_{1}=r_{0}, \quad \varphi_{1}=\varphi_{0} \tag{3.1661} \end{equation*}(3.1661)t1=αs1,r1=r0,φ1=φ0
The value of α α alpha\alphaα can be determined through the requirement that the 4 -velocity has norm one, i.e.,
(3.1662) 1 = ( 1 r r 1 ) t ˙ 2 = ( 1 r r 0 ) α 2 α = 1 1 r r 1 (3.1662) 1 = 1 r r 1 t ˙ 2 = 1 r r 0 α 2 α = 1 1 r r 1 {:(3.1662)1=(1-(r_(**))/(r_(1)))t^(˙)^(2)=(1-(r_(**))/(r_(0)))alpha^(2)quad Longrightarrowquad alpha=(1)/(sqrt(1-(r_(**))/(r_(1)))):}\begin{equation*} 1=\left(1-\frac{r_{*}}{r_{1}}\right) \dot{t}^{2}=\left(1-\frac{r_{*}}{r_{0}}\right) \alpha^{2} \quad \Longrightarrow \quad \alpha=\frac{1}{\sqrt{1-\frac{r_{*}}{r_{1}}}} \tag{3.1662} \end{equation*}(3.1662)1=(1rr1)t˙2=(1rr0)α2α=11rr1
The 4-acceleration is defined as A = V V A = V V A=grad_(V)VA=\nabla_{V} VA=VV. For the worldline of (1), we have V = α t V = α t V=alphadel_(t)V=\alpha \partial_{t}V=αt, and therefore,
(3.1663) A 1 = α 2 t t = α 2 Γ t t μ μ = α 2 r ( r 0 r ) 2 r 0 3 r = r 2 r 0 2 r , (3.1663) A 1 = α 2 t t = α 2 Γ t t μ μ = α 2 r r 0 r 2 r 0 3 r = r 2 r 0 2 r , {:(3.1663)A_(1)=alpha^(2)grad_(t)del_(t)=alpha^(2)Gamma_(tt)^(mu)del_(mu)=alpha^(2)(r_(**)(r_(0)-r_(**)))/(2r_(0)^(3))del_(r)=(r_(**))/(2r_(0)^(2))del_(r)",":}\begin{equation*} A_{1}=\alpha^{2} \nabla_{t} \partial_{t}=\alpha^{2} \Gamma_{t t}^{\mu} \partial_{\mu}=\alpha^{2} \frac{r_{*}\left(r_{0}-r_{*}\right)}{2 r_{0}^{3}} \partial_{r}=\frac{r_{*}}{2 r_{0}^{2}} \partial_{r}, \tag{3.1663} \end{equation*}(3.1663)A1=α2tt=α2Γttμμ=α2r(r0r)2r03r=r2r02r,
where we have used the expression for Γ t t r Γ t t r Gamma_(tt)^(r)\Gamma_{t t}^{r}Γttr, which is the only nonzero Christoffel symbol on the form Γ t t μ Γ t t μ Gamma_(tt)^(mu)\Gamma_{t t}^{\mu}Γttμ. The proper acceleration a a aaa is given by a 2 = g ( A , A ) = a 2 = g ( A , A ) = -a^(2)=g(A,A)=-a^{2}=g(A, A)=a2=g(A,A)= A 2 A 2 A^(2)A^{2}A2, and therefore,
(3.1664) a 1 2 = A 1 2 = ( r 2 r 0 2 ) 2 g r r = r 2 4 r 0 4 ( 1 r / r 0 ) a 1 = r 2 r 0 2 1 r / r 0 (3.1664) a 1 2 = A 1 2 = r 2 r 0 2 2 g r r = r 2 4 r 0 4 1 r / r 0 a 1 = r 2 r 0 2 1 r / r 0 {:(3.1664)a_(1)^(2)=-A_(1)^(2)=-((r_(**))/(2r_(0)^(2)))^(2)g_(rr)=(r_(**)^(2))/(4r_(0)^(4)(1-r_(**)//r_(0)))quad Longrightarrowquada_(1)=(r_(**))/(2r_(0)^(2)sqrt(1-r_(**)//r_(0))):}\begin{equation*} a_{1}^{2}=-A_{1}^{2}=-\left(\frac{r_{*}}{2 r_{0}^{2}}\right)^{2} g_{r r}=\frac{r_{*}^{2}}{4 r_{0}^{4}\left(1-r_{*} / r_{0}\right)} \quad \Longrightarrow \quad a_{1}=\frac{r_{*}}{2 r_{0}^{2} \sqrt{1-r_{*} / r_{0}}} \tag{3.1664} \end{equation*}(3.1664)a12=A12=(r2r02)2grr=r24r04(1r/r0)a1=r2r021r/r0
We can note that, for r 0 r r 0 r r_(0)≫r_(**)r_{0} \gg r_{*}r0r and with r = 2 M G r = 2 M G r_(**)=2MGr_{*}=2 M Gr=2MG, this becomes
(3.1665) a 1 2 M G 2 r 0 2 = M G r 0 2 (3.1665) a 1 2 M G 2 r 0 2 = M G r 0 2 {:(3.1665)a_(1)≃(2MG)/(2r_(0)^(2))=(MG)/(r_(0)^(2)):}\begin{equation*} a_{1} \simeq \frac{2 M G}{2 r_{0}^{2}}=\frac{M G}{r_{0}^{2}} \tag{3.1665} \end{equation*}(3.1665)a12MG2r02=MGr02
which is the expected result from Newtonian gravity.
For the circularly orbiting observer (2), the worldline can be parametrized as
(3.1666) t 2 = β s 2 , r 2 = r 0 , φ 2 = γ s 2 (3.1666) t 2 = β s 2 , r 2 = r 0 , φ 2 = γ s 2 {:(3.1666)t_(2)=betas_(2)","quadr_(2)=r_(0)","quadvarphi_(2)=gammas_(2):}\begin{equation*} t_{2}=\beta s_{2}, \quad r_{2}=r_{0}, \quad \varphi_{2}=\gamma s_{2} \tag{3.1666} \end{equation*}(3.1666)t2=βs2,r2=r0,φ2=γs2
where β β beta\betaβ and γ γ gamma\gammaγ are constants and s 2 s 2 s_(2)s_{2}s2 is the proper time of the observer. By definition, this observer has zero 4-acceleration a = 0 a = 0 a=0a=0a=0, which can be used to fix the constants. The 4-velocity takes the form V 2 = β t + γ φ V 2 = β t + γ φ V_(2)=betadel_(t)+gammadel_(varphi)V_{2}=\beta \partial_{t}+\gamma \partial_{\varphi}V2=βt+γφ and the 4-acceleration is therefore given by
A 2 = V 2 V 2 = β 2 t t + β γ ( t φ + φ t ) + γ 2 φ φ (3.1667) = [ β 2 Γ t t μ + 2 β γ Γ t φ μ + γ 2 Γ φ φ μ ] μ = 0 A 2 = V 2 V 2 = β 2 t t + β γ t φ + φ t + γ 2 φ φ (3.1667) = β 2 Γ t t μ + 2 β γ Γ t φ μ + γ 2 Γ φ φ μ μ = 0 {:[A_(2)=grad_(V_(2))V_(2)=beta^(2)grad_(t)del_(t)+beta gamma(grad_(t)del_(varphi)+grad_(varphi)del_(t))+gamma^(2)grad_(varphi)del_(varphi)],[(3.1667)=[beta^(2)Gamma_(tt)^(mu)+2beta gammaGamma_(t varphi)^(mu)+gamma^(2)Gamma_(varphi varphi)^(mu)]del_(mu)=0]:}\begin{align*} A_{2} & =\nabla_{V_{2}} V_{2}=\beta^{2} \nabla_{t} \partial_{t}+\beta \gamma\left(\nabla_{t} \partial_{\varphi}+\nabla_{\varphi} \partial_{t}\right)+\gamma^{2} \nabla_{\varphi} \partial_{\varphi} \\ & =\left[\beta^{2} \Gamma_{t t}^{\mu}+2 \beta \gamma \Gamma_{t \varphi}^{\mu}+\gamma^{2} \Gamma_{\varphi \varphi}^{\mu}\right] \partial_{\mu}=0 \tag{3.1667} \end{align*}A2=V2V2=β2tt+βγ(tφ+φt)+γ2φφ(3.1667)=[β2Γttμ+2βγΓtφμ+γ2Γφφμ]μ=0
The entire 4 -acceleration being zero requires that each component is zero separately. However, all of the components except the r r rrr-component are trivially vanishing and we are left with
(3.1668) β 2 Γ t t r + 2 β γ Γ t φ r + γ 2 Γ φ φ r = β 2 r ( r r ) 2 r 0 3 γ 2 ( r r ) = 0 (3.1668) β 2 Γ t t r + 2 β γ Γ t φ r + γ 2 Γ φ φ r = β 2 r r r 2 r 0 3 γ 2 r r = 0 {:(3.1668)beta^(2)Gamma_(tt)^(r)+2beta gammaGamma_(t varphi)^(r)+gamma^(2)Gamma_(varphi varphi)^(r)=beta^(2)(r_(**)(r-r_(**)))/(2r_(0)^(3))-gamma^(2)(r-r_(**))=0:}\begin{equation*} \beta^{2} \Gamma_{t t}^{r}+2 \beta \gamma \Gamma_{t \varphi}^{r}+\gamma^{2} \Gamma_{\varphi \varphi}^{r}=\beta^{2} \frac{r_{*}\left(r-r_{*}\right)}{2 r_{0}^{3}}-\gamma^{2}\left(r-r_{*}\right)=0 \tag{3.1668} \end{equation*}(3.1668)β2Γttr+2βγΓtφr+γ2Γφφr=β2r(rr)2r03γ2(rr)=0
leading to
(3.1669) γ = ± β r 2 r 0 3 (3.1669) γ = ± β r 2 r 0 3 {:(3.1669)gamma=+-betasqrt((r_(**))/(2r_(0)^(3))):}\begin{equation*} \gamma= \pm \beta \sqrt{\frac{r_{*}}{2 r_{0}^{3}}} \tag{3.1669} \end{equation*}(3.1669)γ=±βr2r03
where the ± ± +-\pm± just defines the direction of the orbit and we will pick the positive sign in the future. The normalization is again set by the requirement that V 2 = 1 V 2 = 1 V^(2)=1V^{2}=1V2=1, leading to
(3.1670) 1 = ( 1 r r 0 ) β 2 r 0 2 γ 2 = β 2 ( 1 r r 0 r 0 2 r 2 r 0 3 ) = β 2 ( 1 3 r 2 r 0 ) (3.1670) 1 = 1 r r 0 β 2 r 0 2 γ 2 = β 2 1 r r 0 r 0 2 r 2 r 0 3 = β 2 1 3 r 2 r 0 {:(3.1670)1=(1-(r_(**))/(r_(0)))beta^(2)-r_(0)^(2)gamma^(2)=beta^(2)(1-(r_(**))/(r_(0))-r_(0)^(2)(r_(**))/(2r_(0)^(3)))=beta^(2)(1-(3r_(**))/(2r_(0))):}\begin{equation*} 1=\left(1-\frac{r_{*}}{r_{0}}\right) \beta^{2}-r_{0}^{2} \gamma^{2}=\beta^{2}\left(1-\frac{r_{*}}{r_{0}}-r_{0}^{2} \frac{r_{*}}{2 r_{0}^{3}}\right)=\beta^{2}\left(1-\frac{3 r_{*}}{2 r_{0}}\right) \tag{3.1670} \end{equation*}(3.1670)1=(1rr0)β2r02γ2=β2(1rr0r02r2r03)=β2(13r2r0)
and therefore, we find that
(3.1671) β = 1 1 3 r / 2 r 0 (3.1671) β = 1 1 3 r / 2 r 0 {:(3.1671)beta=(1)/(sqrt(1-3r_(**)//2r_(0))):}\begin{equation*} \beta=\frac{1}{\sqrt{1-3 r_{*} / 2 r_{0}}} \tag{3.1671} \end{equation*}(3.1671)β=113r/2r0
The parametrizations of the worldlines are therefore
(3.1672) t 1 = s 1 1 r / r 0 , r 1 = r 0 , φ 1 = φ 0 (3.1672) t 1 = s 1 1 r / r 0 , r 1 = r 0 , φ 1 = φ 0 {:(3.1672)t_(1)=(s_(1))/(sqrt(1-r_(**)//r_(0)))","quadr_(1)=r_(0)","quadvarphi_(1)=varphi_(0):}\begin{equation*} t_{1}=\frac{s_{1}}{\sqrt{1-r_{*} / r_{0}}}, \quad r_{1}=r_{0}, \quad \varphi_{1}=\varphi_{0} \tag{3.1672} \end{equation*}(3.1672)t1=s11r/r0,r1=r0,φ1=φ0
and
(3.1673) t 2 = s 2 1 3 r / 2 r 0 , r 2 = r 0 , φ 2 = s 2 / r 0 2 r 0 / r 3 (3.1673) t 2 = s 2 1 3 r / 2 r 0 , r 2 = r 0 , φ 2 = s 2 / r 0 2 r 0 / r 3 {:(3.1673)t_(2)=(s_(2))/(sqrt(1-3r_(**)//2r_(0)))","quadr_(2)=r_(0)","quadvarphi_(2)=(s_(2)//r_(0))/(sqrt(2r_(0)//r_(**)-3)):}\begin{equation*} t_{2}=\frac{s_{2}}{\sqrt{1-3 r_{*} / 2 r_{0}}}, \quad r_{2}=r_{0}, \quad \varphi_{2}=\frac{s_{2} / r_{0}}{\sqrt{2 r_{0} / r_{*}-3}} \tag{3.1673} \end{equation*}(3.1673)t2=s213r/2r0,r2=r0,φ2=s2/r02r0/r3
respectively, while the proper accelerations are given by
(3.1674) a 1 = r 2 r 0 2 1 r / r 0 and a 2 = 0 (3.1674) a 1 = r 2 r 0 2 1 r / r 0  and  a 2 = 0 {:(3.1674)a_(1)=(r_(**))/(2r_(0)^(2)sqrt(1-r_(**)//r_(0)))quad" and "quada_(2)=0:}\begin{equation*} a_{1}=\frac{r_{*}}{2 r_{0}^{2} \sqrt{1-r_{*} / r_{0}}} \quad \text { and } \quad a_{2}=0 \tag{3.1674} \end{equation*}(3.1674)a1=r2r021r/r0 and a2=0
b) The differences in the proper times will both be proportional to the difference in the coordinate time t t ttt according to
(3.1675) Δ s 1 = Δ t α and Δ s 2 = Δ t β (3.1675) Δ s 1 = Δ t α  and  Δ s 2 = Δ t β {:(3.1675)Deltas_(1)=(Delta t)/(alpha)quad" and "quad Deltas_(2)=(Delta t)/(beta):}\begin{equation*} \Delta s_{1}=\frac{\Delta t}{\alpha} \quad \text { and } \quad \Delta s_{2}=\frac{\Delta t}{\beta} \tag{3.1675} \end{equation*}(3.1675)Δs1=Δtα and Δs2=Δtβ
Therefore, we find that
(3.1676) Δ s 1 Δ s 2 = β α = r 0 r r 0 3 r / 2 ρ . (3.1676) Δ s 1 Δ s 2 = β α = r 0 r r 0 3 r / 2 ρ . {:(3.1676)(Deltas_(1))/(Deltas_(2))=(beta )/(alpha)=sqrt((r_(0)-r_(**))/(r_(0)-3r_(**)//2))-=rho.:}\begin{equation*} \frac{\Delta s_{1}}{\Delta s_{2}}=\frac{\beta}{\alpha}=\sqrt{\frac{r_{0}-r_{*}}{r_{0}-3 r_{*} / 2}} \equiv \rho . \tag{3.1676} \end{equation*}(3.1676)Δs1Δs2=βα=r0rr03r/2ρ.
In particular, we can see that this expression has the correct limits as ρ 1 ρ 1 rho rarr1\rho \rightarrow 1ρ1 when r 0 r 0 r_(0)rarr oor_{0} \rightarrow \inftyr0 and ρ ρ rho rarr oo\rho \rightarrow \inftyρ as r 0 3 r / 2 r 0 3 r / 2 r_(0)rarr3r_(**)//2r_{0} \rightarrow 3 r_{*} / 2r03r/2. The latter result is expected as the circular geodesic at r 0 = 3 r / 2 r 0 = 3 r / 2 r_(0)=3r_(**)//2r_{0}=3 r_{*} / 2r0=3r/2 is lightlike.
c) The relative gamma factor γ ~ = 1 / 1 v 2 γ ~ = 1 / 1 v 2 tilde(gamma)=1//sqrt(1-v^(2))\tilde{\gamma}=1 / \sqrt{1-v^{2}}γ~=1/1v2, where v v vvv is the relative speed, is given by the inner product between the two 4 -velocities, i.e.,
(3.1677) γ ~ = g ( V 1 , V 2 ) = ( 1 r r 0 ) α β = 2 ( r 0 r ) 2 r 0 3 r 2 ( 1 x ) 2 3 x (3.1677) γ ~ = g V 1 , V 2 = 1 r r 0 α β = 2 r 0 r 2 r 0 3 r 2 ( 1 x ) 2 3 x {:(3.1677) tilde(gamma)=g(V_(1),V_(2))=(1-(r_(**))/(r_(0)))alpha beta=sqrt((2(r_(0)-r_(**)))/(2r_(0)-3r_(**)))-=sqrt((2(1-x))/(2-3x)):}\begin{equation*} \tilde{\gamma}=g\left(V_{1}, V_{2}\right)=\left(1-\frac{r_{*}}{r_{0}}\right) \alpha \beta=\sqrt{\frac{2\left(r_{0}-r_{*}\right)}{2 r_{0}-3 r_{*}}} \equiv \sqrt{\frac{2(1-x)}{2-3 x}} \tag{3.1677} \end{equation*}(3.1677)γ~=g(V1,V2)=(1rr0)αβ=2(r0r)2r03r2(1x)23x
where x = r / r 0 x = r / r 0 x=r_(**)//r_(0)x=r_{*} / r_{0}x=r/r0. Squaring this relation leads to
(3.1678) 1 v 2 = 2 3 x 2 ( 1 x ) v = x 2 ( 1 x ) = r 2 ( r 0 r ) (3.1678) 1 v 2 = 2 3 x 2 ( 1 x ) v = x 2 ( 1 x ) = r 2 r 0 r {:(3.1678)1-v^(2)=(2-3x)/(2(1-x))quad Longrightarrowquad v=sqrt((x)/(2(1-x)))=sqrt((r_(**))/(2(r_(0)-r_(**)))):}\begin{equation*} 1-v^{2}=\frac{2-3 x}{2(1-x)} \quad \Longrightarrow \quad v=\sqrt{\frac{x}{2(1-x)}}=\sqrt{\frac{r_{*}}{2\left(r_{0}-r_{*}\right)}} \tag{3.1678} \end{equation*}(3.1678)1v2=23x2(1x)v=x2(1x)=r2(r0r)
Note that we again recover the classical orbital velocity
(3.1679) v = M G r 0 (3.1679) v = M G r 0 {:(3.1679)v=sqrt((MG)/(r_(0))):}\begin{equation*} v=\sqrt{\frac{M G}{r_{0}}} \tag{3.1679} \end{equation*}(3.1679)v=MGr0
in the limit where r 0 r = 2 M G r 0 r = 2 M G r_(0)≫r_(**)=2MGr_{0} \gg r_{*}=2 M Gr0r=2MG.

2.101

a) The proper time τ τ tau\tauτ can be found from
(3.1680) d τ 2 = d s 2 (3.1680) d τ 2 = d s 2 {:(3.1680)dtau^(2)=ds^(2):}\begin{equation*} d \tau^{2}=d s^{2} \tag{3.1680} \end{equation*}(3.1680)dτ2=ds2
Using the Kerr metric with d r = 0 d r = 0 dr=0d r=0dr=0 and d θ = 0 d θ = 0 d theta=0d \theta=0dθ=0 (since r = R r = R r=Rr=Rr=R and θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2 ) for the given orbit, we get
d τ 2 = ( 1 r R ) d t 2 + 2 a r R d t d ϕ ( R 2 + a 2 + a 2 r R ) d ϕ 2 = [ ( 1 r R ) + 2 a r R d ϕ d t ( R 2 + a 2 + a 2 r R ) ( d ϕ d t ) 2 ] d t 2 (3.1681) = [ ( 1 r R ) + 2 a r R 2 v ( 1 + a 2 R 2 + a 2 r R 3 ) v 2 ] d t 2 d τ 2 = 1 r R d t 2 + 2 a r R d t d ϕ R 2 + a 2 + a 2 r R d ϕ 2 = 1 r R + 2 a r R d ϕ d t R 2 + a 2 + a 2 r R d ϕ d t 2 d t 2 (3.1681) = 1 r R + 2 a r R 2 v 1 + a 2 R 2 + a 2 r R 3 v 2 d t 2 {:[dtau^(2)=(1-(r_(**))/(R))dt^(2)+(2ar_(**))/(R)dtd phi-(R^(2)+a^(2)+(a^(2)r_(**))/(R))dphi^(2)],[=[(1-(r_(**))/(R))+(2ar_(**))/(R)(d phi)/(dt)-(R^(2)+a^(2)+(a^(2)r_(**))/(R))((d phi)/(dt))^(2)]dt^(2)],[(3.1681)=[(1-(r_(**))/(R))+(2ar_(**))/(R^(2))v-(1+(a^(2))/(R^(2))+(a^(2)r_(**))/(R^(3)))v^(2)]dt^(2)]:}\begin{align*} d \tau^{2} & =\left(1-\frac{r_{*}}{R}\right) d t^{2}+\frac{2 a r_{*}}{R} d t d \phi-\left(R^{2}+a^{2}+\frac{a^{2} r_{*}}{R}\right) d \phi^{2} \\ & =\left[\left(1-\frac{r_{*}}{R}\right)+\frac{2 a r_{*}}{R} \frac{d \phi}{d t}-\left(R^{2}+a^{2}+\frac{a^{2} r_{*}}{R}\right)\left(\frac{d \phi}{d t}\right)^{2}\right] d t^{2} \\ & =\left[\left(1-\frac{r_{*}}{R}\right)+\frac{2 a r_{*}}{R^{2}} v-\left(1+\frac{a^{2}}{R^{2}}+\frac{a^{2} r_{*}}{R^{3}}\right) v^{2}\right] d t^{2} \tag{3.1681} \end{align*}dτ2=(1rR)dt2+2arRdtdϕ(R2+a2+a2rR)dϕ2=[(1rR)+2arRdϕdt(R2+a2+a2rR)(dϕdt)2]dt2(3.1681)=[(1rR)+2arR2v(1+a2R2+a2rR3)v2]dt2
since ρ 2 = R 2 ρ 2 = R 2 rho^(2)=R^(2)\rho^{2}=R^{2}ρ2=R2 and v = R d ϕ / d t v = R d ϕ / d t v=Rd phi//dtv=R d \phi / d tv=Rdϕ/dt. This implies that
(3.1682) d τ = ± ( 1 r R ) + 2 a r R 2 v ( 1 + a 2 R 2 + a 2 r R 3 ) v 2 d t (3.1682) d τ = ± 1 r R + 2 a r R 2 v 1 + a 2 R 2 + a 2 r R 3 v 2 d t {:(3.1682)d tau=+-sqrt((1-(r_(**))/(R))+(2ar_(**))/(R^(2))v-(1+(a^(2))/(R^(2))+(a^(2)r_(**))/(R^(3)))v^(2))dt:}\begin{equation*} d \tau= \pm \sqrt{\left(1-\frac{r_{*}}{R}\right)+\frac{2 a r_{*}}{R^{2}} v-\left(1+\frac{a^{2}}{R^{2}}+\frac{a^{2} r_{*}}{R^{3}}\right) v^{2}} d t \tag{3.1682} \end{equation*}(3.1682)dτ=±(1rR)+2arR2v(1+a2R2+a2rR3)v2dt
where the square root is independent of time t t ttt. Thus, using d t / d τ > 0 d t / d τ > 0 dt//d tau > 0d t / d \tau>0dt/dτ>0 and integrating from t = t 0 = 0 t = t 0 = 0 t=t_(0)=0t=t_{0}=0t=t0=0 to t = T = 2 π R / v t = T = 2 π R / v t=T=2pi R//vt=T=2 \pi R / vt=T=2πR/v, we obtain the proper time
(3.1683) τ = T ( 1 r R ) + 2 a r R 2 v ( 1 + a 2 R 2 + a 2 r R 3 ) v 2 (3.1683) τ = T 1 r R + 2 a r R 2 v 1 + a 2 R 2 + a 2 r R 3 v 2 {:(3.1683)tau=Tsqrt((1-(r_(**))/(R))+(2ar_(**))/(R^(2))v-(1+(a^(2))/(R^(2))+(a^(2)r_(**))/(R^(3)))v^(2)):}\begin{equation*} \tau=T \sqrt{\left(1-\frac{r_{*}}{R}\right)+\frac{2 a r_{*}}{R^{2}} v-\left(1+\frac{a^{2}}{R^{2}}+\frac{a^{2} r_{*}}{R^{3}}\right) v^{2}} \tag{3.1683} \end{equation*}(3.1683)τ=T(1rR)+2arR2v(1+a2R2+a2rR3)v2
b) Solving the result in a) for T / τ T / τ T//tauT / \tauT/τ and series expanding in v v vvv and r / R r / R r_(**)//Rr_{*} / Rr/R up to orders v 2 v 2 v^(2)v^{2}v2 and r / R r / R r_(**)//Rr_{*} / Rr/R (assuming v 2 r / R v 2 r / R v^(2)∼r_(**)//Rv^{2} \sim r_{*} / Rv2r/R ), we obtain
T τ = 1 ( 1 r R ) + 2 a r R 2 v ( 1 + a 2 R 2 + a 2 r R 3 ) v 2 (3.1684) 1 + 1 2 ( 1 + a 2 R 2 ) v 2 + 1 2 R r T τ = 1 1 r R + 2 a r R 2 v 1 + a 2 R 2 + a 2 r R 3 v 2 (3.1684) 1 + 1 2 1 + a 2 R 2 v 2 + 1 2 R r {:[(T)/( tau)=(1)/(sqrt((1-(r_(**))/(R))+(2ar_(**))/(R^(2))v-(1+(a^(2))/(R^(2))+(a^(2)r_(**))/(R^(3)))v^(2)))],[(3.1684)≃1+(1)/(2)(1+(a^(2))/(R^(2)))v^(2)+(1)/(2R)r_(**)]:}\begin{align*} \frac{T}{\tau} & =\frac{1}{\sqrt{\left(1-\frac{r_{*}}{R}\right)+\frac{2 a r_{*}}{R^{2}} v-\left(1+\frac{a^{2}}{R^{2}}+\frac{a^{2} r_{*}}{R^{3}}\right) v^{2}}} \\ & \simeq 1+\frac{1}{2}\left(1+\frac{a^{2}}{R^{2}}\right) v^{2}+\frac{1}{2 R} r_{*} \tag{3.1684} \end{align*}Tτ=1(1rR)+2arR2v(1+a2R2+a2rR3)v2(3.1684)1+12(1+a2R2)v2+12Rr
2.102
a) If we assume a fixed time t = t 0 t = t 0 t=t_(0)t=t_{0}t=t0, we observe that the metric is just a flat metric, i.e.,
(3.1685) d s 2 = a ( t 0 ) 2 e 2 C t 0 ( d x 2 + d y 2 + d z 2 ) (3.1685) d s 2 = a t 0 2 e 2 C t 0 d x 2 + d y 2 + d z 2 {:(3.1685)ds^(2)=-a(t_(0))^(2)e^(2Ct_(0))(dx^(2)+dy^(2)+dz^(2)):}\begin{equation*} d s^{2}=-a\left(t_{0}\right)^{2} e^{2 C t_{0}}\left(d x^{2}+d y^{2}+d z^{2}\right) \tag{3.1685} \end{equation*}(3.1685)ds2=a(t0)2e2Ct0(dx2+dy2+dz2)
where x = ρ sin θ cos ϕ , y = ρ sin θ sin ϕ x = ρ sin θ cos ϕ , y = ρ sin θ sin ϕ x=rho sin theta cos phi,y=rho sin theta sin phix=\rho \sin \theta \cos \phi, y=\rho \sin \theta \sin \phix=ρsinθcosϕ,y=ρsinθsinϕ, and z = ρ cos θ z = ρ cos θ z=rho cos thetaz=\rho \cos \thetaz=ρcosθ are expressed in the spherical coordinates ρ , θ ρ , θ rho,theta\rho, \thetaρ,θ, and ϕ ϕ phi\phiϕ. Note that a ( t 0 ) e C t 0 a t 0 e C t 0 a(t_(0))e^(Ct_(0))a\left(t_{0}\right) e^{C t_{0}}a(t0)eCt0 is a constant, since t 0 t 0 t_(0)t_{0}t0 is fixed and C C CCC is a constant.
b) We choose to use the Cartesian coordinates x , y x , y x,yx, yx,y, and z z zzz. In order to find a geodesic in the spacetime, we use the Lagrangian
(3.1686) L = t ˙ 2 a ( t ) 2 ( x ˙ 2 + y ˙ 2 + z ˙ 2 ) = 1 (3.1686) L = t ˙ 2 a ( t ) 2 x ˙ 2 + y ˙ 2 + z ˙ 2 = 1 {:(3.1686)L=t^(˙)^(2)-a(t)^(2)(x^(˙)^(2)+y^(˙)^(2)+z^(˙)^(2))=1:}\begin{equation*} \mathcal{L}=\dot{t}^{2}-a(t)^{2}\left(\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2}\right)=1 \tag{3.1686} \end{equation*}(3.1686)L=t˙2a(t)2(x˙2+y˙2+z˙2)=1
where dot means differentiation with respect to proper time τ τ tau\tauτ. Now, the metric does not depend on x , y x , y x,yx, yx,y, and z z zzz, so using the Euler-Lagrange equations d d τ ( L x ˙ ) d d τ L x ˙ (d)/(d tau)((delL)/(delx^(˙)^(')))-\frac{d}{d \tau}\left(\frac{\partial \mathcal{L}}{\partial \dot{x}^{\prime}}\right)-ddτ(Lx˙) L x i = 0 L x i = 0 (delL)/(delx^(i))=0\frac{\partial \mathcal{L}}{\partial x^{i}}=0Lxi=0, we have three constants of motion, which are given as follows
(3.1687) d d τ ( a ( t ) 2 x ˙ i ) = 0 a ( t ) 2 x ˙ i = k i (3.1687) d d τ a ( t ) 2 x ˙ i = 0 a ( t ) 2 x ˙ i = k i {:(3.1687)(d)/(d tau)(a(t)^(2)x^(˙)^(i))=0quad=>quad a(t)^(2)x^(˙)^(i)=k_(i):}\begin{equation*} \frac{d}{d \tau}\left(a(t)^{2} \dot{x}^{i}\right)=0 \quad \Rightarrow \quad a(t)^{2} \dot{x}^{i}=k_{i} \tag{3.1687} \end{equation*}(3.1687)ddτ(a(t)2x˙i)=0a(t)2x˙i=ki
where k i , i = 1 , 2 , 3 k i , i = 1 , 2 , 3 k_(i),i=1,2,3k_{i}, i=1,2,3ki,i=1,2,3, are the constants of motion. Inserting the three constants of motion into the Lagrangian, we obtain
(3.1688) t ˙ 2 1 a ( t ) 2 ( k 1 2 + k 2 2 + k 3 2 ) = 1 , (3.1688) t ˙ 2 1 a ( t ) 2 k 1 2 + k 2 2 + k 3 2 = 1 , {:(3.1688)t^(˙)^(2)-(1)/(a(t)^(2))(k_(1)^(2)+k_(2)^(2)+k_(3)^(2))=1",":}\begin{equation*} \dot{t}^{2}-\frac{1}{a(t)^{2}}\left(k_{1}^{2}+k_{2}^{2}+k_{3}^{2}\right)=1, \tag{3.1688} \end{equation*}(3.1688)t˙21a(t)2(k12+k22+k32)=1,
where a ( t ) = a 0 e C t a ( t ) = a 0 e C t a(t)=a_(0)e^(Ct)a(t)=a_{0} e^{C t}a(t)=a0eCt. Then, rewrite the initial conditions, we find
d x d τ | τ = 0 = ( sin θ cos ϕ d ρ d τ + ρ cos θ cos ϕ d θ d τ ρ sin θ sin ϕ d ϕ d τ ) | τ = 0 (3.1689) = sin 0 cos 0 1 4 + 1 cos 0 cos 0 1 4 1 sin 0 sin 0 1 4 = 1 4 d y d τ | τ = 0 = ( sin θ sin ϕ d ρ d τ + ρ cos θ sin ϕ d θ d τ + ρ sin θ cos ϕ d ϕ d τ ) | τ = 0 (3.1690) = sin 0 sin 0 1 4 + 1 cos 0 sin 0 1 4 + 1 sin 0 cos 0 1 4 = 0 , (3.1691) d z d τ | τ = 0 = ( cos θ d ρ d τ ρ sin θ d θ d τ ) | τ = 0 = cos 0 1 4 1 sin 0 1 4 = 1 4 , d x d τ τ = 0 = sin θ cos ϕ d ρ d τ + ρ cos θ cos ϕ d θ d τ ρ sin θ sin ϕ d ϕ d τ τ = 0 (3.1689) = sin 0 cos 0 1 4 + 1 cos 0 cos 0 1 4 1 sin 0 sin 0 1 4 = 1 4 d y d τ τ = 0 = sin θ sin ϕ d ρ d τ + ρ cos θ sin ϕ d θ d τ + ρ sin θ cos ϕ d ϕ d τ τ = 0 (3.1690) = sin 0 sin 0 1 4 + 1 cos 0 sin 0 1 4 + 1 sin 0 cos 0 1 4 = 0 , (3.1691) d z d τ τ = 0 = cos θ d ρ d τ ρ sin θ d θ d τ τ = 0 = cos 0 1 4 1 sin 0 1 4 = 1 4 , {:[(dx)/(d tau)|_(tau=0)=(sin theta cos phi(d rho)/(d tau)+rho cos theta cos phi(d theta)/(d tau)-rho sin theta sin phi(d phi)/(d tau))|_(tau=0)],[(3.1689)=sin 0*cos 0*(1)/(4)+1*cos 0*cos 0*(1)/(4)-1*sin 0*sin 0*(1)/(4)=(1)/(4)],[(dy)/(d tau)|_(tau=0)=(sin theta sin phi(d rho)/(d tau)+rho cos theta sin phi(d theta)/(d tau)+rho sin theta cos phi(d phi)/(d tau))|_(tau=0)],[(3.1690)=sin 0*sin 0*(1)/(4)+1*cos 0*sin 0*(1)/(4)+1*sin 0*cos 0*(1)/(4)=0","],[(3.1691)(dz)/(d tau)|_(tau=0)=(cos theta(d rho)/(d tau)-rho sin theta(d theta)/(d tau))|_(tau=0)=cos 0*(1)/(4)-1*sin 0*(1)/(4)=(1)/(4)","]:}\begin{align*} \left.\frac{d x}{d \tau}\right|_{\tau=0} & =\left.\left(\sin \theta \cos \phi \frac{d \rho}{d \tau}+\rho \cos \theta \cos \phi \frac{d \theta}{d \tau}-\rho \sin \theta \sin \phi \frac{d \phi}{d \tau}\right)\right|_{\tau=0} \\ & =\sin 0 \cdot \cos 0 \cdot \frac{1}{4}+1 \cdot \cos 0 \cdot \cos 0 \cdot \frac{1}{4}-1 \cdot \sin 0 \cdot \sin 0 \cdot \frac{1}{4}=\frac{1}{4} \tag{3.1689}\\ \left.\frac{d y}{d \tau}\right|_{\tau=0} & =\left.\left(\sin \theta \sin \phi \frac{d \rho}{d \tau}+\rho \cos \theta \sin \phi \frac{d \theta}{d \tau}+\rho \sin \theta \cos \phi \frac{d \phi}{d \tau}\right)\right|_{\tau=0} \\ & =\sin 0 \cdot \sin 0 \cdot \frac{1}{4}+1 \cdot \cos 0 \cdot \sin 0 \cdot \frac{1}{4}+1 \cdot \sin 0 \cdot \cos 0 \cdot \frac{1}{4}=0, \tag{3.1690}\\ \left.\frac{d z}{d \tau}\right|_{\tau=0} & =\left.\left(\cos \theta \frac{d \rho}{d \tau}-\rho \sin \theta \frac{d \theta}{d \tau}\right)\right|_{\tau=0}=\cos 0 \cdot \frac{1}{4}-1 \cdot \sin 0 \cdot \frac{1}{4}=\frac{1}{4}, \tag{3.1691} \end{align*}dxdτ|τ=0=(sinθcosϕdρdτ+ρcosθcosϕdθdτρsinθsinϕdϕdτ)|τ=0(3.1689)=sin0cos014+1cos0cos0141sin0sin014=14dydτ|τ=0=(sinθsinϕdρdτ+ρcosθsinϕdθdτ+ρsinθcosϕdϕdτ)|τ=0(3.1690)=sin0sin014+1cos0sin014+1sin0cos014=0,(3.1691)dzdτ|τ=0=(cosθdρdτρsinθdθdτ)|τ=0=cos0141sin014=14,
which lead to
(3.1692) a ( t ) 2 x ˙ | τ = 0 = a 0 2 4 = k 1 , a ( t ) 2 y ˙ | τ = 0 = 0 = k 2 , a ( t ) 2 z ˙ | τ = 0 = a 0 2 4 = k 3 (3.1692) a ( t ) 2 x ˙ τ = 0 = a 0 2 4 = k 1 , a ( t ) 2 y ˙ τ = 0 = 0 = k 2 , a ( t ) 2 z ˙ τ = 0 = a 0 2 4 = k 3 {:(3.1692)a(t)^(2)(x^(˙))|_(tau=0)=(a_(0)^(2))/(4)=k_(1)"," quad a(t)^(2)(y^(˙))|_(tau=0)=0=k_(2)"," quad a(t)^(2)(z^(˙))|_(tau=0)=(a_(0)^(2))/(4)=k_(3):}\begin{equation*} \left.a(t)^{2} \dot{x}\right|_{\tau=0}=\frac{a_{0}^{2}}{4}=k_{1},\left.\quad a(t)^{2} \dot{y}\right|_{\tau=0}=0=k_{2},\left.\quad a(t)^{2} \dot{z}\right|_{\tau=0}=\frac{a_{0}^{2}}{4}=k_{3} \tag{3.1692} \end{equation*}(3.1692)a(t)2x˙|τ=0=a024=k1,a(t)2y˙|τ=0=0=k2,a(t)2z˙|τ=0=a024=k3
since t | τ = 0 = 0 t τ = 0 = 0 t|_(tau=0)=0\left.t\right|_{\tau=0}=0t|τ=0=0 and a ( 0 ) = 0 a ( 0 ) = 0 a(0)=0a(0)=0a(0)=0, which yield
(3.1693) k 1 = k 3 = a 0 2 4 , k 2 = 0 (3.1693) k 1 = k 3 = a 0 2 4 , k 2 = 0 {:(3.1693)k_(1)=k_(3)=(a_(0)^(2))/(4)","quadk_(2)=0:}\begin{equation*} k_{1}=k_{3}=\frac{a_{0}^{2}}{4}, \quad k_{2}=0 \tag{3.1693} \end{equation*}(3.1693)k1=k3=a024,k2=0
and imply that k 1 2 + k 2 2 + k 3 2 = a 0 4 / 8 k 1 2 + k 2 2 + k 3 2 = a 0 4 / 8 k_(1)^(2)+k_(2)^(2)+k_(3)^(2)=a_(0)^(4)//8k_{1}^{2}+k_{2}^{2}+k_{3}^{2}=a_{0}^{4} / 8k12+k22+k32=a04/8. Thus, we obtain
(3.1694) t ˙ 2 1 a 0 2 e 2 C t a 0 4 8 = 1 t ˙ 2 = 1 + a 0 2 8 e 2 C t d t d τ = ± 1 + a 0 2 8 e 2 C t (3.1694) t ˙ 2 1 a 0 2 e 2 C t a 0 4 8 = 1 t ˙ 2 = 1 + a 0 2 8 e 2 C t d t d τ = ± 1 + a 0 2 8 e 2 C t {:(3.1694)t^(˙)^(2)-(1)/(a_(0)^(2)e^(2Ct))(a_(0)^(4))/(8)=1quad=>quadt^(˙)^(2)=1+(a_(0)^(2))/(8)e^(-2Ct)quad=>quad(dt)/(d tau)=+-sqrt(1+(a_(0)^(2))/(8)e^(-2Ct)):}\begin{equation*} \dot{t}^{2}-\frac{1}{a_{0}^{2} e^{2 C t}} \frac{a_{0}^{4}}{8}=1 \quad \Rightarrow \quad \dot{t}^{2}=1+\frac{a_{0}^{2}}{8} e^{-2 C t} \quad \Rightarrow \quad \frac{d t}{d \tau}= \pm \sqrt{1+\frac{a_{0}^{2}}{8} e^{-2 C t}} \tag{3.1694} \end{equation*}(3.1694)t˙21a02e2Cta048=1t˙2=1+a028e2Ctdtdτ=±1+a028e2Ct
which can be separated to
(3.1695) d τ = ± d t 1 + a 0 2 8 e 2 C t (3.1695) d τ = ± d t 1 + a 0 2 8 e 2 C t {:(3.1695)d tau=+-(dt)/(sqrt(1+(a_(0)^(2))/(8)e^(-2Ct))):}\begin{equation*} d \tau= \pm \frac{d t}{\sqrt{1+\frac{a_{0}^{2}}{8} e^{-2 C t}}} \tag{3.1695} \end{equation*}(3.1695)dτ=±dt1+a028e2Ct
Finally, using t ˙ = d t / d τ > 0 t ˙ = d t / d τ > 0 t^(˙)=dt//d tau > 0\dot{t}=d t / d \tau>0t˙=dt/dτ>0, we calculate the proper time Δ τ Δ τ Delta tau\Delta \tauΔτ for the free-falling particle between the coordinate times t = 0 t = 0 t=0t=0t=0 and t = t 1 t = t 1 t=t_(1)t=t_{1}t=t1
(3.1696) Δ τ = 0 t 1 d t 1 + a 0 2 8 e 2 C t = 1 C ( artanh 1 1 + a 0 2 8 e 2 C t 1 artanh 1 1 + a 0 2 8 ) (3.1696) Δ τ = 0 t 1 d t 1 + a 0 2 8 e 2 C t = 1 C artanh 1 1 + a 0 2 8 e 2 C t 1 artanh 1 1 + a 0 2 8 {:(3.1696)Delta tau=int_(0)^(t_(1))(dt)/(sqrt(1+(a_(0)^(2))/(8)e^(-2Ct)))=(1)/(C)(artanh((1)/(sqrt(1+(a_(0)^(2))/(8)*e^(-2Ct_(1)))))-artanh((1)/(sqrt(1+(a_(0)^(2))/(8))))):}\begin{equation*} \Delta \tau=\int_{0}^{t_{1}} \frac{d t}{\sqrt{1+\frac{a_{0}^{2}}{8} e^{-2 C t}}}=\frac{1}{C}\left(\operatorname{artanh} \frac{1}{\sqrt{1+\frac{a_{0}^{2}}{8} \cdot e^{-2 C t_{1}}}}-\operatorname{artanh} \frac{1}{\sqrt{1+\frac{a_{0}^{2}}{8}}}\right) \tag{3.1696} \end{equation*}(3.1696)Δτ=0t1dt1+a028e2Ct=1C(artanh11+a028e2Ct1artanh11+a028)

2.103

a) Let the initial and final times be t = 0 t = 0 t=0t=0t=0 and t = T F t = T F t=T_(F)t=T_{F}t=TF, respectively. Assuming that v = a ( t ) R 0 d ϕ d t = v = a ( t ) R 0 d ϕ d t = v=a(t)R_(0)(d phi)/(dt)=v=a(t) R_{0} \frac{d \phi}{d t}=v=a(t)R0dϕdt= const. and using
(3.1697) d τ = 1 a ( t ) 2 R 0 2 ( d ϕ d t ) 2 d t = 1 v 2 d t (3.1697) d τ = 1 a ( t ) 2 R 0 2 d ϕ d t 2 d t = 1 v 2 d t {:(3.1697)d tau=sqrt(1-a(t)^(2)R_(0)^(2)((d phi)/(dt))^(2))dt=sqrt(1-v^(2))dt:}\begin{equation*} d \tau=\sqrt{1-a(t)^{2} R_{0}^{2}\left(\frac{d \phi}{d t}\right)^{2}} d t=\sqrt{1-v^{2}} d t \tag{3.1697} \end{equation*}(3.1697)dτ=1a(t)2R02(dϕdt)2dt=1v2dt
we obtain the proper time from t = 0 t = 0 t=0t=0t=0 to t = T F t = T F t=T_(F)t=T_{F}t=TF as
(3.1698) τ = 0 T F 1 v 2 d t = 1 v 2 T F (3.1698) τ = 0 T F 1 v 2 d t = 1 v 2 T F {:(3.1698)tau=int_(0)^(T_(F))sqrt(1-v^(2))dt=sqrt(1-v^(2))T_(F):}\begin{equation*} \tau=\int_{0}^{T_{F}} \sqrt{1-v^{2}} d t=\sqrt{1-v^{2}} T_{F} \tag{3.1698} \end{equation*}(3.1698)τ=0TF1v2dt=1v2TF
Now, we need to find T F T F T_(F)T_{F}TF. From the fact that v v vvv is a constant and using v = e t R 0 d ϕ d t v = e t R 0 d ϕ d t v=e^(t)R_(0)(d phi)/(dt)v=e^{t} R_{0} \frac{d \phi}{d t}v=etR0dϕdt, we can calculate how ϕ ϕ phi\phiϕ depends on t t ttt :
(3.1699) e t d t = R 0 v d ϕ ( e T F 1 ) = R 0 v 2 π (3.1699) e t d t = R 0 v d ϕ e T F 1 = R 0 v 2 π {:(3.1699)e^(-t)dt=(R_(0))/(v)d phiquad=>quad-(e^(-T_(F))-1)=(R_(0))/(v)2pi:}\begin{equation*} e^{-t} d t=\frac{R_{0}}{v} d \phi \quad \Rightarrow \quad-\left(e^{-T_{F}}-1\right)=\frac{R_{0}}{v} 2 \pi \tag{3.1699} \end{equation*}(3.1699)etdt=R0vdϕ(eTF1)=R0v2π
which implies that
(3.1700) T F = ln ( 1 2 π R 0 v ) 1 (3.1700) T F = ln 1 2 π R 0 v 1 {:(3.1700)T_(F)=ln (1-(2piR_(0))/(v))^(-1):}\begin{equation*} T_{F}=\ln \left(1-\frac{2 \pi R_{0}}{v}\right)^{-1} \tag{3.1700} \end{equation*}(3.1700)TF=ln(12πR0v)1
b) No, T F T F T_(F)T_{F}TF becomes infinite if 1 2 π R 0 / v = 0 1 2 π R 0 / v = 0 1-2piR_(0)//v=01-2 \pi R_{0} / v=012πR0/v=0.

2.104

The Robertson-Walker metric with zero curvature is
(3.1701) d s 2 = d t 2 a ( t ) 2 ( d x 2 + d y 2 + d z 2 ) (3.1701) d s 2 = d t 2 a ( t ) 2 d x 2 + d y 2 + d z 2 {:(3.1701)ds^(2)=dt^(2)-a(t)^(2)(dx^(2)+dy^(2)+dz^(2)):}\begin{equation*} d s^{2}=d t^{2}-a(t)^{2}\left(d x^{2}+d y^{2}+d z^{2}\right) \tag{3.1701} \end{equation*}(3.1701)ds2=dt2a(t)2(dx2+dy2+dz2)
The Lagrangian is given by
(3.1702) L = t ˙ 2 a ( t ) 2 ( x ˙ 2 + y ˙ 2 + z ˙ 2 ) = 1 (3.1702) L = t ˙ 2 a ( t ) 2 x ˙ 2 + y ˙ 2 + z ˙ 2 = 1 {:(3.1702)L=t^(˙)^(2)-a(t)^(2)(x^(˙)^(2)+y^(˙)^(2)+z^(˙)^(2))=1:}\begin{equation*} \mathcal{L}=\dot{t}^{2}-a(t)^{2}\left(\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2}\right)=1 \tag{3.1702} \end{equation*}(3.1702)L=t˙2a(t)2(x˙2+y˙2+z˙2)=1
The Euler-Lagrange equation with respect to x x xxx leads to
(3.1703) d d τ ( a ( t ) 2 x ˙ ) = 0 a ( t ) 2 x ˙ = c 1 (3.1703) d d τ a ( t ) 2 x ˙ = 0 a ( t ) 2 x ˙ = c 1 {:(3.1703)(d)/(d tau)(a(t)^(2)(x^(˙)))=0quad=>quad a(t)^(2)x^(˙)=c_(1):}\begin{equation*} \frac{d}{d \tau}\left(a(t)^{2} \dot{x}\right)=0 \quad \Rightarrow \quad a(t)^{2} \dot{x}=c_{1} \tag{3.1703} \end{equation*}(3.1703)ddτ(a(t)2x˙)=0a(t)2x˙=c1
and together with the initial conditions x ˙ ( τ = 0 ) = A x ˙ ( τ = 0 ) = A x^(˙)(tau=0)=A\dot{x}(\tau=0)=Ax˙(τ=0)=A and t ( τ = 0 ) = 0 t ( τ = 0 ) = 0 t(tau=0)=0t(\tau=0)=0t(τ=0)=0, we obtain c 1 = a ( 0 ) 2 A c 1 = a ( 0 ) 2 A c_(1)=a(0)^(2)Ac_{1}=a(0)^{2} Ac1=a(0)2A, and thus, in total, we find that
(3.1704) x ˙ = a ( 0 ) 2 A a ( t ) 2 (3.1704) x ˙ = a ( 0 ) 2 A a ( t ) 2 {:(3.1704)x^(˙)=(a(0)^(2)A)/(a(t)^(2)):}\begin{equation*} \dot{x}=\frac{a(0)^{2} A}{a(t)^{2}} \tag{3.1704} \end{equation*}(3.1704)x˙=a(0)2Aa(t)2
Similarly, using the same procedure, we can conclude that y ˙ = z ˙ = 0 y ˙ = z ˙ = 0 y^(˙)=z^(˙)=0\dot{y}=\dot{z}=0y˙=z˙=0. In principle, we can determine t t ttt as a function of τ τ tau\tauτ from the Lagrangian
(3.1705) L = t ˙ 2 a ( 0 ) 4 A 2 a ( t ) 2 = 1 t ˙ 2 = 1 + a ( 0 ) 4 A 2 a ( t ) 2 i = 1 + a ( 0 ) 4 A 2 a ( t ) 2 (3.1705) L = t ˙ 2 a ( 0 ) 4 A 2 a ( t ) 2 = 1 t ˙ 2 = 1 + a ( 0 ) 4 A 2 a ( t ) 2 i = 1 + a ( 0 ) 4 A 2 a ( t ) 2 {:(3.1705)L=t^(˙)^(2)-(a(0)^(4)A^(2))/(a(t)^(2))=1=>t^(˙)^(2)=1+(a(0)^(4)A^(2))/(a(t)^(2))quad=>quad i=sqrt(1+(a(0)^(4)A^(2))/(a(t)^(2))):}\begin{equation*} \mathcal{L}=\dot{t}^{2}-\frac{a(0)^{4} A^{2}}{a(t)^{2}}=1 \Rightarrow \dot{t}^{2}=1+\frac{a(0)^{4} A^{2}}{a(t)^{2}} \quad \Rightarrow \quad i=\sqrt{1+\frac{a(0)^{4} A^{2}}{a(t)^{2}}} \tag{3.1705} \end{equation*}(3.1705)L=t˙2a(0)4A2a(t)2=1t˙2=1+a(0)4A2a(t)2i=1+a(0)4A2a(t)2
The equation x ˙ = a ( 0 ) 2 A / a ( t ) 2 x ˙ = a ( 0 ) 2 A / a ( t ) 2 x^(˙)=a(0)^(2)A//a(t)^(2)\dot{x}=a(0)^{2} A / a(t)^{2}x˙=a(0)2A/a(t)2 can now be rearranged as
(3.1706) a ( t ) 2 a ( 0 ) 2 A d x = d τ (3.1706) a ( t ) 2 a ( 0 ) 2 A d x = d τ {:(3.1706)(a(t)^(2))/(a(0)^(2)A)dx=d tau:}\begin{equation*} \frac{a(t)^{2}}{a(0)^{2} A} d x=d \tau \tag{3.1706} \end{equation*}(3.1706)a(t)2a(0)2Adx=dτ
Integrating both sides, we obtain
(3.1707) τ = X 0 X D a ( t ) 2 a ( 0 ) 2 A d x (3.1707) τ = X 0 X D a ( t ) 2 a ( 0 ) 2 A d x {:(3.1707)tau=int_(X_(0))^(X_(D))(a(t)^(2))/(a(0)^(2)A)dx:}\begin{equation*} \tau=\int_{X_{0}}^{X_{D}} \frac{a(t)^{2}}{a(0)^{2} A} d x \tag{3.1707} \end{equation*}(3.1707)τ=X0XDa(t)2a(0)2Adx

2.105

a) The Schwarzschild metric is given by
(3.1708) d s 2 = ( 1 r r ) d t 2 ( 1 r r ) 1 d r 2 r 2 d Ω 2 (3.1708) d s 2 = 1 r r d t 2 1 r r 1 d r 2 r 2 d Ω 2 {:(3.1708)ds^(2)=(1-(r_(**))/(r))dt^(2)-(1-(r_(**))/(r))^(-1)dr^(2)-r^(2)dOmega^(2):}\begin{equation*} d s^{2}=\left(1-\frac{r_{*}}{r}\right) d t^{2}-\left(1-\frac{r_{*}}{r}\right)^{-1} d r^{2}-r^{2} d \Omega^{2} \tag{3.1708} \end{equation*}(3.1708)ds2=(1rr)dt2(1rr)1dr2r2dΩ2
See Figure 3.15 for the setup of observers A A AAA and B B BBB. We will assume motion in the plane θ = π 2 θ = π 2 theta=(pi)/(2)\theta=\frac{\pi}{2}θ=π2 and so d θ = 0 d θ = 0 d theta=0d \theta=0dθ=0 and d Ω 2 = d φ 2 d Ω 2 = d φ 2 dOmega^(2)=dvarphi^(2)d \Omega^{2}=d \varphi^{2}dΩ2=dφ2. From the symmetries of the spacetime (Killing vector fields t t del_(t)\partial_{t}t and φ φ del_(varphi)\partial_{\varphi}φ ), we have the conserved quantities
(3.1709) E = g ( t , γ ˙ ) = ( 1 r r ) t ˙ , L = g ( φ , γ ˙ ) = r 2 φ ˙ (3.1709) E = g t , γ ˙ = 1 r r t ˙ , L = g φ , γ ˙ = r 2 φ ˙ {:(3.1709)E=g(del_(t),(gamma^(˙)))=(1-(r_(**))/(r))t^(˙)","quad L=-g(del_(varphi),(gamma^(˙)))=r^(2)varphi^(˙):}\begin{equation*} E=g\left(\partial_{t}, \dot{\gamma}\right)=\left(1-\frac{r_{*}}{r}\right) \dot{t}, \quad L=-g\left(\partial_{\varphi}, \dot{\gamma}\right)=r^{2} \dot{\varphi} \tag{3.1709} \end{equation*}(3.1709)E=g(t,γ˙)=(1rr)t˙,L=g(φ,γ˙)=r2φ˙
Figure 3.15 Setup of the black hole ("BH") and observers A A AAA and B B BBB.
Observer A A AAA is not in geodesic motion but is stationary. The worldline of A A AAA can therefore be parametrized by
(3.1710) t = α 0 s , r = r 0 (3.1710) t = α 0 s , r = r 0 {:(3.1710)t=alpha_(0)s","quad r=r_(0):}\begin{equation*} t=\alpha_{0} s, \quad r=r_{0} \tag{3.1710} \end{equation*}(3.1710)t=α0s,r=r0
where s s sss is the proper time. Normalizing this such that V A = t ˙ t + r ˙ r = α 0 t V A = t ˙ t + r ˙ r = α 0 t V_(A)=t^(˙)del_(t)+r^(˙)del_(r)=alpha_(0)del_(t)V_{A}=\dot{t} \partial_{t}+\dot{r} \partial_{r}=\alpha_{0} \partial_{t}VA=t˙t+r˙r=α0t satisfies V A 2 = 1 V A 2 = 1 V_(A)^(2)=1V_{A}^{2}=1VA2=1, we find that
(3.1711) α 2 g t t = α 0 2 ( 1 r r 0 ) = 1 α 0 = 1 1 r / r 0 (3.1711) α 2 g t t = α 0 2 1 r r 0 = 1 α 0 = 1 1 r / r 0 {:(3.1711)alpha^(2)g_(tt)=alpha_(0)^(2)(1-(r_(**))/(r_(0)))=1quad=>quadalpha_(0)=(1)/(sqrt(1-r_(**)//r_(0))):}\begin{equation*} \alpha^{2} g_{t t}=\alpha_{0}^{2}\left(1-\frac{r_{*}}{r_{0}}\right)=1 \quad \Rightarrow \quad \alpha_{0}=\frac{1}{\sqrt{1-r_{*} / r_{0}}} \tag{3.1711} \end{equation*}(3.1711)α2gtt=α02(1rr0)=1α0=11r/r0
For observer B B BBB, the 4-velocity is purely radial and at r = r 1 r = r 1 r=r_(1)r=r_{1}r=r1 is given by V B 0 = α 1 t V B 0 = α 1 t V_(B0)=alpha_(1)del_(t)V_{B 0}=\alpha_{1} \partial_{t}VB0=α1t. Normalization of V B 0 V B 0 V_(B0)V_{B 0}VB0 leads to
(3.1712) α 1 2 g ( t , t ) = α 1 2 ( 1 r r 1 ) = 1 α 1 = 1 1 r / r 1 (3.1712) α 1 2 g t , t = α 1 2 1 r r 1 = 1 α 1 = 1 1 r / r 1 {:(3.1712)alpha_(1)^(2)g(del_(t),del_(t))=alpha_(1)^(2)(1-(r_(**))/(r_(1)))=1quad=>quadalpha_(1)=(1)/(sqrt(1-r_(**)//r_(1))):}\begin{equation*} \alpha_{1}^{2} g\left(\partial_{t}, \partial_{t}\right)=\alpha_{1}^{2}\left(1-\frac{r_{*}}{r_{1}}\right)=1 \quad \Rightarrow \quad \alpha_{1}=\frac{1}{\sqrt{1-r_{*} / r_{1}}} \tag{3.1712} \end{equation*}(3.1712)α12g(t,t)=α12(1rr1)=1α1=11r/r1
At radius r r rrr, the 4 -velocity of B B BBB is given by
(3.1713) V B = α 1 ( r ) t β ( r ) r = t ˙ t + r ˙ r (3.1713) V B = α 1 ( r ) t β ( r ) r = t ˙ t + r ˙ r {:(3.1713)V_(B)=alpha_(1)(r)del_(t)-beta(r)del_(r)=t^(˙)del_(t)+r^(˙)del_(r):}\begin{equation*} V_{B}=\alpha_{1}(r) \partial_{t}-\beta(r) \partial_{r}=\dot{t} \partial_{t}+\dot{r} \partial_{r} \tag{3.1713} \end{equation*}(3.1713)VB=α1(r)tβ(r)r=t˙t+r˙r
Since B B BBB is freely falling and t t del_(t)\partial_{t}t is a Killing vector field, it follows that
(3.1714) g ( V B , t ) = g t t α 1 ( r ) = ( 1 r r ) α 1 ( r ) = E , (3.1714) g V B , t = g t t α 1 ( r ) = 1 r r α 1 ( r ) = E , {:(3.1714)g(V_(B),del_(t))=g_(tt)alpha_(1)(r)=(1-(r_(**))/(r))alpha_(1)(r)=E",":}\begin{equation*} g\left(V_{B}, \partial_{t}\right)=g_{t t} \alpha_{1}(r)=\left(1-\frac{r_{*}}{r}\right) \alpha_{1}(r)=E, \tag{3.1714} \end{equation*}(3.1714)g(VB,t)=gttα1(r)=(1rr)α1(r)=E,
is a constant of motion. From r = r 1 r = r 1 r=r_(1)r=r_{1}r=r1, we find
(3.1715) E = 1 r r 1 1 r / r 1 = 1 r / r 1 α 1 ( r ) = 1 r / r 1 1 r r (3.1715) E = 1 r r 1 1 r / r 1 = 1 r / r 1 α 1 ( r ) = 1 r / r 1 1 r r {:(3.1715)E=(1-(r_(**))/(r_(1)))/(sqrt(1-r_(**)//r_(1)))=sqrt(1-r_(**)//r_(1))=>alpha_(1)(r)=(sqrt(1-r_(**)//r_(1)))/(1-(r_(**))/(r)):}\begin{equation*} E=\frac{1-\frac{r_{*}}{r_{1}}}{\sqrt{1-r_{*} / r_{1}}}=\sqrt{1-r_{*} / r_{1}} \Rightarrow \alpha_{1}(r)=\frac{\sqrt{1-r_{*} / r_{1}}}{1-\frac{r_{*}}{r}} \tag{3.1715} \end{equation*}(3.1715)E=1rr11r/r1=1r/r1α1(r)=1r/r11rr
The relative γ γ gamma\gammaγ factor between A A AAA and B B BBB as they pass each other is given by
γ = g ( V A , V B ) = α 0 α 1 ( r 0 ) g t t = 1 1 r / r 0 1 r / r 1 1 r r 0 ( 1 r r 0 ) (3.1716) = 1 r / r 1 1 r / r 0 1 1 v 2 , γ = g V A , V B = α 0 α 1 r 0 g t t = 1 1 r / r 0 1 r / r 1 1 r r 0 1 r r 0 (3.1716) = 1 r / r 1 1 r / r 0 1 1 v 2 , {:[gamma=g(V_(A),V_(B))=alpha_(0)alpha_(1)(r_(0))g_(tt)=(1)/(sqrt(1-r_(**)//r_(0)))(sqrt(1-r_(**)//r_(1)))/(1-(r_(**))/(r_(0)))(1-(r_(**))/(r_(0)))],[(3.1716)=sqrt((1-r_(**)//r_(1))/(1-r_(**)//r_(0)))-=(1)/(sqrt(1-v^(2)))","]:}\begin{align*} \gamma & =g\left(V_{A}, V_{B}\right)=\alpha_{0} \alpha_{1}\left(r_{0}\right) g_{t t}=\frac{1}{\sqrt{1-r_{*} / r_{0}}} \frac{\sqrt{1-r_{*} / r_{1}}}{1-\frac{r_{*}}{r_{0}}}\left(1-\frac{r_{*}}{r_{0}}\right) \\ & =\sqrt{\frac{1-r_{*} / r_{1}}{1-r_{*} / r_{0}}} \equiv \frac{1}{\sqrt{1-v^{2}}}, \tag{3.1716} \end{align*}γ=g(VA,VB)=α0α1(r0)gtt=11r/r01r/r11rr0(1rr0)(3.1716)=1r/r11r/r011v2,
where v v vvv is the relative velocity between A A AAA and B B BBB by definition. Solving for v v vvv, we obtain
v = 1 1 γ 2 = 1 1 r / r 0 1 r / r 1 = 1 1 r / r 1 1 r r 1 1 + r r 0 (3.1717) = r 1 r / r 1 1 r 0 1 r 1 . v = 1 1 γ 2 = 1 1 r / r 0 1 r / r 1 = 1 1 r / r 1 1 r r 1 1 + r r 0 (3.1717) = r 1 r / r 1 1 r 0 1 r 1 . {:[v=sqrt(1-(1)/(gamma^(2)))=sqrt(1-(1-r_(**)//r_(0))/(1-r_(**)//r_(1)))=(1)/(sqrt(1-r_(**)//r_(1)))sqrt(1-(r_(**))/(r_(1))-1+(r_(**))/(r_(0)))],[(3.1717)=sqrt((r_(**))/(1-r_(**)//r_(1)))sqrt((1)/(r_(0))-(1)/(r_(1))).]:}\begin{align*} v & =\sqrt{1-\frac{1}{\gamma^{2}}}=\sqrt{1-\frac{1-r_{*} / r_{0}}{1-r_{*} / r_{1}}}=\frac{1}{\sqrt{1-r_{*} / r_{1}}} \sqrt{1-\frac{r_{*}}{r_{1}}-1+\frac{r_{*}}{r_{0}}} \\ & =\sqrt{\frac{r_{*}}{1-r_{*} / r_{1}}} \sqrt{\frac{1}{r_{0}}-\frac{1}{r_{1}}} . \tag{3.1717} \end{align*}v=11γ2=11r/r01r/r1=11r/r11rr11+rr0(3.1717)=r1r/r11r01r1.
Note that, while β ( r ) β ( r ) beta(r)\beta(r)β(r) can be computed from the normalization of V B V B V_(B)V_{B}VB, it is not necessary to find the relative velocity.
b) From the normalization of B B BBB 's 4 -velocity, we find that
( 1 r r ) t ˙ 2 ( 1 r r ) 1 r ˙ 2 = ( 1 r r ) 1 E 2 ( 1 r r ) 1 r ˙ 2 = 1 (3.1718) E 2 r ˙ 2 = 1 r r , 1 r r t ˙ 2 1 r r 1 r ˙ 2 = 1 r r 1 E 2 1 r r 1 r ˙ 2 = 1 (3.1718) E 2 r ˙ 2 = 1 r r , {:[(1-(r_(**))/(r))t^(˙)^(2)-(1-(r_(**))/(r))^(-1)r^(˙)^(2)=(1-(r_(**))/(r))^(-1)E^(2)-(1-(r_(**))/(r))^(-1)r^(˙)^(2)=1],[(3.1718)=>quadE^(2)-r^(˙)^(2)=1-(r_(**))/(r)","]:}\begin{align*} & \left(1-\frac{r_{*}}{r}\right) \dot{t}^{2}-\left(1-\frac{r_{*}}{r}\right)^{-1} \dot{r}^{2}=\left(1-\frac{r_{*}}{r}\right)^{-1} E^{2}-\left(1-\frac{r_{*}}{r}\right)^{-1} \dot{r}^{2}=1 \\ & \Rightarrow \quad E^{2}-\dot{r}^{2}=1-\frac{r_{*}}{r}, \tag{3.1718} \end{align*}(1rr)t˙2(1rr)1r˙2=(1rr)1E2(1rr)1r˙2=1(3.1718)E2r˙2=1rr,
where r ˙ = d r d τ r ˙ = d r d τ r^(˙)=(dr)/(d tau)\dot{r}=\frac{d r}{d \tau}r˙=drdτ. Solving for r ˙ r ˙ r^(˙)\dot{r}r˙ leads to
(3.1719) r ˙ = E 2 1 + r r = r r r r 1 . (3.1719) r ˙ = E 2 1 + r r = r r r r 1 . {:(3.1719)r^(˙)=-sqrt(E^(2)-1+(r_(**))/(r))=-sqrt((r_(**))/(r)-(r_(**))/(r_(1))).:}\begin{equation*} \dot{r}=-\sqrt{E^{2}-1+\frac{r_{*}}{r}}=-\sqrt{\frac{r_{*}}{r}-\frac{r_{*}}{r_{1}}} . \tag{3.1719} \end{equation*}(3.1719)r˙=E21+rr=rrrr1.
This is a separable differential equation, which integrates to
(3.1720) τ = r 1 r 0 d r r r r r 1 = r 0 r 1 r r 1 r ( r 1 r ) d r (3.1720) τ = r 1 r 0 d r r r r r 1 = r 0 r 1 r r 1 r r 1 r d r {:(3.1720)tau=-int_(r_(1))^(r_(0))(dr)/(sqrt((r_(**))/(r)-(r_(**))/(r_(1))))=int_(r_(0))^(r_(1))sqrt((rr_(1))/(r_(**)(r_(1)-r)))dr:}\begin{equation*} \tau=-\int_{r_{1}}^{r_{0}} \frac{d r}{\sqrt{\frac{r_{*}}{r}-\frac{r_{*}}{r_{1}}}}=\int_{r_{0}}^{r_{1}} \sqrt{\frac{r r_{1}}{r_{*}\left(r_{1}-r\right)}} d r \tag{3.1720} \end{equation*}(3.1720)τ=r1r0drrrrr1=r0r1rr1r(r1r)dr
which is therefore the proper time for B B BBB between r 1 r 1 r_(1)r_{1}r1 and r 0 r 0 r_(0)r_{0}r0.

2.106

a) For an observer with x = x 0 x = x 0 x=x_(0)x=x_{0}x=x0, we have d x = 0 d x = 0 dx=0d x=0dx=0, and therefore, we find that
(3.1721) d s 2 = x 0 2 d t 2 d t d s = + 1 x 0 t = s x 0 + t 0 (3.1721) d s 2 = x 0 2 d t 2 d t d s = + 1 x 0 t = s x 0 + t 0 {:(3.1721)ds^(2)=x_(0)^(2)dt^(2)quad=>quad(dt)/(ds)=+(1)/(x_(0))quad=>quad t=(s)/(x_(0))+t_(0):}\begin{equation*} d s^{2}=x_{0}^{2} d t^{2} \quad \Rightarrow \quad \frac{d t}{d s}=+\frac{1}{x_{0}} \quad \Rightarrow \quad t=\frac{s}{x_{0}}+t_{0} \tag{3.1721} \end{equation*}(3.1721)ds2=x02dt2dtds=+1x0t=sx0+t0
The worldline can therefore be described by (choosing t 0 = 0 t 0 = 0 t_(0)=0t_{0}=0t0=0 )
(3.1722) x = x 0 , t = s x 0 , (3.1722) x = x 0 , t = s x 0 , {:(3.1722)x=x_(0)","quad t=(s)/(x_(0))",":}\begin{equation*} x=x_{0}, \quad t=\frac{s}{x_{0}}, \tag{3.1722} \end{equation*}(3.1722)x=x0,t=sx0,
where s s sss is the proper time. The 4-acceleration A A AAA is defined by A = γ ˙ γ ˙ = ( χ ¨ μ + A = γ ˙ γ ˙ = χ ¨ μ + A=grad_(gamma^(˙))gamma^(˙)=(chi^(¨)^(mu)+:}A=\nabla_{\dot{\gamma}} \dot{\gamma}=\left(\ddot{\chi}^{\mu}+\right.A=γ˙γ˙=(χ¨μ+ Γ ν σ μ χ ˙ v χ ˙ σ ) μ Γ ν σ μ χ ˙ v χ ˙ σ μ {:Gamma_(nu sigma)^(mu)chi^(˙)^(v)chi^(˙)^(sigma))del_(mu)\left.\Gamma_{\nu \sigma}^{\mu} \dot{\chi}^{v} \dot{\chi}^{\sigma}\right) \partial_{\mu}Γνσμχ˙vχ˙σ)μ. For our worldline, we have x ˙ = 0 , t ˙ = 1 x 0 x ˙ = 0 , t ˙ = 1 x 0 x^(˙)=0,t^(˙)=(1)/(x_(0))\dot{x}=0, \dot{t}=\frac{1}{x_{0}}x˙=0,t˙=1x0, and x ¨ = t ¨ = 0 x ¨ = t ¨ = 0 x^(¨)=t^(¨)=0\ddot{x}=\ddot{t}=0x¨=t¨=0, and therefore, we obtain
(3.1723) A = Γ t t a x 0 2 a (3.1723) A = Γ t t a x 0 2 a {:(3.1723)A=Gamma_(tt)^(a)x_(0)^(-2)del_(a):}\begin{equation*} A=\Gamma_{t t}^{a} x_{0}^{-2} \partial_{a} \tag{3.1723} \end{equation*}(3.1723)A=Γttax02a
We know that
Γ t t a = 1 2 g a b ( t g t b + t g t b b g t t ) (3.1724) = 1 2 g a b ( 0 + 0 b g t t ) = 1 2 g a b b g t t = 1 2 g a b b ( x 2 ) Γ t t a = 1 2 g a b t g t b + t g t b b g t t (3.1724) = 1 2 g a b 0 + 0 b g t t = 1 2 g a b b g t t = 1 2 g a b b x 2 {:[Gamma_(tt)^(a)=(1)/(2)g^(ab)(del_(t)g_(tb)+del_(t)g_(tb)-del_(b)g_(tt))],[(3.1724)=(1)/(2)g^(ab)(0+0-del_(b)g_(tt))=-(1)/(2)g^(ab)del_(b)g_(tt)=-(1)/(2)g^(ab)del_(b)(x^(2))]:}\begin{align*} \Gamma_{t t}^{a} & =\frac{1}{2} g^{a b}\left(\partial_{t} g_{t b}+\partial_{t} g_{t b}-\partial_{b} g_{t t}\right) \\ & =\frac{1}{2} g^{a b}\left(0+0-\partial_{b} g_{t t}\right)=-\frac{1}{2} g^{a b} \partial_{b} g_{t t}=-\frac{1}{2} g^{a b} \partial_{b}\left(x^{2}\right) \tag{3.1724} \end{align*}Γtta=12gab(tgtb+tgtbbgtt)(3.1724)=12gab(0+0bgtt)=12gabbgtt=12gabb(x2)
Thus, we identify Γ t t t = 0 Γ t t t = 0 Gamma_(tt)^(t)=0\Gamma_{t t}^{t}=0Γttt=0 and Γ t t x = 1 2 g x x x x 2 = 1 2 ( 1 ) 2 x = x Γ t t x = 1 2 g x x x x 2 = 1 2 ( 1 ) 2 x = x Gamma_(tt)^(x)=-(1)/(2)g^(xx)del_(x)x^(2)=-(1)/(2)(-1)2x=x\Gamma_{t t}^{x}=-\frac{1}{2} g^{x x} \partial_{x} x^{2}=-\frac{1}{2}(-1) 2 x=xΓttx=12gxxxx2=12(1)2x=x. Therefore, for our worldline, we have
(3.1725) A = x 0 1 x (3.1725) A = x 0 1 x {:(3.1725)A=x_(0)^(-1)del_(x):}\begin{equation*} A=x_{0}^{-1} \partial_{x} \tag{3.1725} \end{equation*}(3.1725)A=x01x
The proper acceleration α α alpha\alphaα is given by α 2 = A 2 = g ( A , A ) = x 0 2 g ( x , x ) = α 2 = A 2 = g ( A , A ) = x 0 2 g x , x = alpha^(2)=-A^(2)=-g(A,A)=-x_(0)^(-2)g(del_(x),del_(x))=\alpha^{2}=-A^{2}=-g(A, A)=-x_{0}^{-2} g\left(\partial_{x}, \partial_{x}\right)=α2=A2=g(A,A)=x02g(x,x)= x 0 2 x 0 2 x_(0)^(-2)x_{0}^{-2}x02, which implies that
(3.1726) α = 1 x 0 (3.1726) α = 1 x 0 {:(3.1726)alpha=(1)/(x_(0)):}\begin{equation*} \alpha=\frac{1}{x_{0}} \tag{3.1726} \end{equation*}(3.1726)α=1x0
b) We have the Killing vector field K = t K = t K=del_(t)K=\partial_{t}K=t. Since a free-falling observer moves along a geodesic, we have the constant of motion
(3.1727) Q = g ( t , γ ˙ ) = x 2 t ˙ t ˙ = Q x 2 (3.1727) Q = g t , γ ˙ = x 2 t ˙ t ˙ = Q x 2 {:(3.1727)Q=g(del_(t),(gamma^(˙)))=x^(2)t^(˙)quad=>quadt^(˙)=(Q)/(x^(2)):}\begin{equation*} Q=g\left(\partial_{t}, \dot{\gamma}\right)=x^{2} \dot{t} \quad \Rightarrow \quad \dot{t}=\frac{Q}{x^{2}} \tag{3.1727} \end{equation*}(3.1727)Q=g(t,γ˙)=x2t˙t˙=Qx2
We also know that g ( γ ˙ , γ ˙ ) = x 2 t ˙ 2 x ˙ 2 = Q 2 x 2 x ˙ 2 = 1 g ( γ ˙ , γ ˙ ) = x 2 t ˙ 2 x ˙ 2 = Q 2 x 2 x ˙ 2 = 1 g(gamma^(˙),gamma^(˙))=x^(2)t^(˙)^(2)-x^(˙)^(2)=(Q^(2))/(x^(2))-x^(˙)^(2)=1g(\dot{\gamma}, \dot{\gamma})=x^{2} \dot{t}^{2}-\dot{x}^{2}=\frac{Q^{2}}{x^{2}}-\dot{x}^{2}=1g(γ˙,γ˙)=x2t˙2x˙2=Q2x2x˙2=1. This leads to
(3.1728) x ˙ = Q 2 x 2 1 = d x d s (3.1728) x ˙ = Q 2 x 2 1 = d x d s {:(3.1728)x^(˙)=sqrt((Q^(2))/(x^(2))-1)=(dx)/(ds):}\begin{equation*} \dot{x}=\sqrt{\frac{Q^{2}}{x^{2}}-1}=\frac{d x}{d s} \tag{3.1728} \end{equation*}(3.1728)x˙=Q2x21=dxds
Since we start with d x d t = 0 d x d t = 0 (dx)/(dt)=0\frac{d x}{d t}=0dxdt=0, we know that Q 2 = x 0 2 Q = + x 0 Q 2 = x 0 2 Q = + x 0 Q^(2)=x_(0)^(2)rarr Q=+x_(0)Q^{2}=x_{0}^{2} \rightarrow Q=+x_{0}Q2=x02Q=+x0. To compute the proper time s s sss to reach x = 0 x = 0 x=0x=0x=0, we need to compute the integral
s = d s = 0 x 0 d s d x d x = 0 x 0 d x x 0 2 x 2 1 = 0 x 0 x d x x 0 2 x 2 = { x = x 0 sin θ , d x = x 0 cos θ d θ } = x 0 0 arcsin 1 sin θ 1 sin 2 θ cos θ d θ (3.1729) = x 0 0 arcsin 1 sin θ d θ = x 0 [ cos θ ] 0 arcsin 1 = x 0 s = d s = 0 x 0 d s d x d x = 0 x 0 d x x 0 2 x 2 1 = 0 x 0 x d x x 0 2 x 2 = x = x 0 sin θ , d x = x 0 cos θ d θ = x 0 0 arcsin 1 sin θ 1 sin 2 θ cos θ d θ (3.1729) = x 0 0 arcsin 1 sin θ d θ = x 0 [ cos θ ] 0 arcsin 1 = x 0 {:[s=int ds=int_(0)^(x_(0))(ds)/(dx)dx=int_(0)^(x_(0))(dx)/(sqrt((x_(0)^(2))/(x^(2))-1))],[=int_(0)^(x_(0))(xdx)/(sqrt(x_(0)^(2)-x^(2)))={x=x_(0)sin theta,dx=x_(0)cos theta d theta}],[=x_(0)int_(0)^(arcsin 1)(sin theta)/(sqrt(1-sin^(2)theta))cos theta d theta],[(3.1729)=x_(0)int_(0)^(arcsin 1)sin theta d theta=-x_(0)[cos theta]_(0)^(arcsin 1)=x_(0)]:}\begin{align*} s & =\int d s=\int_{0}^{x_{0}} \frac{d s}{d x} d x=\int_{0}^{x_{0}} \frac{d x}{\sqrt{\frac{x_{0}^{2}}{x^{2}}-1}} \\ & =\int_{0}^{x_{0}} \frac{x d x}{\sqrt{x_{0}^{2}-x^{2}}}=\left\{x=x_{0} \sin \theta, d x=x_{0} \cos \theta d \theta\right\} \\ & =x_{0} \int_{0}^{\arcsin 1} \frac{\sin \theta}{\sqrt{1-\sin ^{2} \theta}} \cos \theta d \theta \\ & =x_{0} \int_{0}^{\arcsin 1} \sin \theta d \theta=-x_{0}[\cos \theta]_{0}^{\arcsin 1}=x_{0} \tag{3.1729} \end{align*}s=ds=0x0dsdxdx=0x0dxx02x21=0x0xdxx02x2={x=x0sinθ,dx=x0cosθdθ}=x00arcsin1sinθ1sin2θcosθdθ(3.1729)=x00arcsin1sinθdθ=x0[cosθ]0arcsin1=x0
Note that the given coordinates are Rindler coordinates on Minkowski space. We can also easily deduce s = x 0 s = x 0 s=x_(0)s=x_{0}s=x0 by transforming to Minkowski coordinates.

2.107

Using the first relation u v = ( 2 μ r ) e ( r 2 μ ) / 2 μ u v = ( 2 μ r ) e ( r 2 μ ) / 2 μ uv=(2mu-r)e^((r-2mu)//2mu)u v=(2 \mu-r) e^{(r-2 \mu) / 2 \mu}uv=(2μr)e(r2μ)/2μ, where u < 0 u < 0 u < 0u<0u<0 and v > 0 v > 0 v > 0v>0v>0, we obtain by differentiation and Leibniz' rule
u d v + v d u = r d r e ( r 2 μ ) / 2 μ + ( 2 μ r ) 1 2 μ e ( r 2 μ ) / 2 μ d r = 1 2 μ e ( r 2 μ ) / 2 μ r d r u d v + v d u = r d r e ( r 2 μ ) / 2 μ + ( 2 μ r ) 1 2 μ e ( r 2 μ ) / 2 μ d r = 1 2 μ e ( r 2 μ ) / 2 μ r d r udv+vdu=-rdre^((r-2mu)//2mu)+(2mu-r)(1)/(2mu)e^((r-2mu)//2mu)dr=-(1)/(2mu)e^((r-2mu)//2mu)rdru d v+v d u=-r d r e^{(r-2 \mu) / 2 \mu}+(2 \mu-r) \frac{1}{2 \mu} e^{(r-2 \mu) / 2 \mu} d r=-\frac{1}{2 \mu} e^{(r-2 \mu) / 2 \mu} r d rudv+vdu=rdre(r2μ)/2μ+(2μr)12μe(r2μ)/2μdr=12μe(r2μ)/2μrdr.
Similarly, using the second relation t = 2 μ ln ( v / u ) t = 2 μ ln ( v / u ) t=2mu ln(-v//u)t=2 \mu \ln (-v / u)t=2μln(v/u), we obtain
(3.1731) 1 2 μ d t = 1 v d v 1 u d u (3.1731) 1 2 μ d t = 1 v d v 1 u d u {:(3.1731)(1)/(2mu)dt=(1)/(v)dv-(1)/(u)du:}\begin{equation*} \frac{1}{2 \mu} d t=\frac{1}{v} d v-\frac{1}{u} d u \tag{3.1731} \end{equation*}(3.1731)12μdt=1vdv1udu
which implies that
(3.1732) u d v v d u = 1 2 μ u v d t = 1 2 μ ( 2 μ r ) e ( r 2 μ ) / 2 μ d t (3.1732) u d v v d u = 1 2 μ u v d t = 1 2 μ ( 2 μ r ) e ( r 2 μ ) / 2 μ d t {:(3.1732)udv-vdu=(1)/(2mu)uvdt=(1)/(2mu)*(2mu-r)e^((r-2mu)//2mu)dt:}\begin{equation*} u d v-v d u=\frac{1}{2 \mu} u v d t=\frac{1}{2 \mu} \cdot(2 \mu-r) e^{(r-2 \mu) / 2 \mu} d t \tag{3.1732} \end{equation*}(3.1732)udvvdu=12μuvdt=12μ(2μr)e(r2μ)/2μdt
Solving for d u d u dud udu and d v d v dvd vdv, we find that
(3.1733) 2 u d v = [ 1 2 μ ( 2 μ r ) d t 1 2 μ r d r ] e ( r 2 μ ) / 2 μ (3.1734) 2 v d u = [ 1 2 μ ( 2 μ r ) d t 1 2 μ r d r ] e ( r 2 μ ) / 2 μ (3.1733) 2 u d v = 1 2 μ ( 2 μ r ) d t 1 2 μ r d r e ( r 2 μ ) / 2 μ (3.1734) 2 v d u = 1 2 μ ( 2 μ r ) d t 1 2 μ r d r e ( r 2 μ ) / 2 μ {:[(3.1733)2udv=[(1)/(2mu)*(2mu-r)dt-(1)/(2mu)rdr]e^((r-2mu)//2mu)],[(3.1734)2vdu=[-(1)/(2mu)(2mu-r)dt-(1)/(2mu)rdr]e^((r-2mu)//2mu)]:}\begin{align*} & 2 u d v=\left[\frac{1}{2 \mu} \cdot(2 \mu-r) d t-\frac{1}{2 \mu} r d r\right] e^{(r-2 \mu) / 2 \mu} \tag{3.1733}\\ & 2 v d u=\left[-\frac{1}{2 \mu}(2 \mu-r) d t-\frac{1}{2 \mu} r d r\right] e^{(r-2 \mu) / 2 \mu} \tag{3.1734} \end{align*}(3.1733)2udv=[12μ(2μr)dt12μrdr]e(r2μ)/2μ(3.1734)2vdu=[12μ(2μr)dt12μrdr]e(r2μ)/2μ
Thus, we have
4 d u d v = 1 u v 2 u d v 2 v d u (3.1735) = 1 2 μ r [ 1 4 μ 2 ( 2 μ r ) 2 d t 2 + 1 4 μ 2 r 2 d r 2 ] e ( r 2 μ ) / 2 μ 4 d u d v = 1 u v 2 u d v 2 v d u (3.1735) = 1 2 μ r 1 4 μ 2 ( 2 μ r ) 2 d t 2 + 1 4 μ 2 r 2 d r 2 e ( r 2 μ ) / 2 μ {:[4dudv=(1)/(uv)2udv*2vdu],[(3.1735)=(1)/(2mu-r)[-(1)/(4mu^(2))(2mu-r)^(2)dt^(2)+(1)/(4mu^(2))r^(2)dr^(2)]e^((r-2mu)//2mu)]:}\begin{align*} 4 d u d v & =\frac{1}{u v} 2 u d v \cdot 2 v d u \\ & =\frac{1}{2 \mu-r}\left[-\frac{1}{4 \mu^{2}}(2 \mu-r)^{2} d t^{2}+\frac{1}{4 \mu^{2}} r^{2} d r^{2}\right] e^{(r-2 \mu) / 2 \mu} \tag{3.1735} \end{align*}4dudv=1uv2udv2vdu(3.1735)=12μr[14μ2(2μr)2dt2+14μ2r2dr2]e(r2μ)/2μ
and finally, assuming μ G M μ G M mu-=GM\mu \equiv G MμGM, we obtain the equivalence between the KruskalSzekeres metric and the standard Schwarzschild metric as
16 μ 2 r e ( r 2 μ ) / 2 μ d u d v = 4 μ 2 r ( 2 μ r ) [ 1 4 μ 2 ( 2 μ r ) 2 d t 2 + 1 4 μ 2 r 2 d r 2 ] = 2 μ r r d t 2 + r 2 μ r d r 2 = ( 1 2 μ r ) d t 2 ( 1 2 μ r ) 1 d r 2 (3.1736) = ( 1 2 G M r ) d x 0 ( 1 2 G M r ) 1 d r 2 16 μ 2 r e ( r 2 μ ) / 2 μ d u d v = 4 μ 2 r ( 2 μ r ) 1 4 μ 2 ( 2 μ r ) 2 d t 2 + 1 4 μ 2 r 2 d r 2 = 2 μ r r d t 2 + r 2 μ r d r 2 = 1 2 μ r d t 2 1 2 μ r 1 d r 2 (3.1736) = 1 2 G M r d x 0 1 2 G M r 1 d r 2 {:[(16mu^(2))/(r)*e^(-(r-2mu)//2mu)dudv=(4mu^(2))/(r(2mu-r))[-(1)/(4mu^(2))(2mu-r)^(2)dt^(2)+(1)/(4mu^(2))r^(2)dr^(2)]],[=-(2mu-r)/(r)dt^(2)+(r)/(2mu-r)dr^(2)],[=(1-(2mu)/(r))dt^(2)-(1-(2mu)/(r))^(-1)dr^(2)],[(3.1736)=(1-(2GM)/(r))dx^(0)-(1-(2GM)/(r))^(-1)dr^(2)]:}\begin{align*} \frac{16 \mu^{2}}{r} \cdot e^{-(r-2 \mu) / 2 \mu} d u d v & =\frac{4 \mu^{2}}{r(2 \mu-r)}\left[-\frac{1}{4 \mu^{2}}(2 \mu-r)^{2} d t^{2}+\frac{1}{4 \mu^{2}} r^{2} d r^{2}\right] \\ & =-\frac{2 \mu-r}{r} d t^{2}+\frac{r}{2 \mu-r} d r^{2} \\ & =\left(1-\frac{2 \mu}{r}\right) d t^{2}-\left(1-\frac{2 \mu}{r}\right)^{-1} d r^{2} \\ & =\left(1-\frac{2 G M}{r}\right) d x^{0}-\left(1-\frac{2 G M}{r}\right)^{-1} d r^{2} \tag{3.1736} \end{align*}16μ2re(r2μ)/2μdudv=4μ2r(2μr)[14μ2(2μr)2dt2+14μ2r2dr2]=2μrrdt2+r2μrdr2=(12μr)dt2(12μr)1dr2(3.1736)=(12GMr)dx0(12GMr)1dr2

2.108

The metric is then given by
(3.1737) d s 2 = 16 μ 2 r e ( 2 μ r ) / 2 μ d u d v r 2 d Ω 2 (3.1737) d s 2 = 16 μ 2 r e ( 2 μ r ) / 2 μ d u d v r 2 d Ω 2 {:(3.1737)ds^(2)=(16mu^(2))/(r)e^((2mu-r)//2mu)dudv-r^(2)dOmega^(2):}\begin{equation*} d s^{2}=\frac{16 \mu^{2}}{r} e^{(2 \mu-r) / 2 \mu} d u d v-r^{2} d \Omega^{2} \tag{3.1737} \end{equation*}(3.1737)ds2=16μ2re(2μr)/2μdudvr2dΩ2
where μ = G M / c 2 μ = G M / c 2 mu=GM//c^(2)\mu=G M / c^{2}μ=GM/c2 and r r rrr (as well as the time t = t ( u , v ) t = t ( u , v ) t=t(u,v)t=t(u, v)t=t(u,v), see below) is a function of u u uuu and v v vvv. The coordinate r r rrr is defined by the equation
(3.1738) u v = ( 2 μ r ) e ( r 2 μ ) / 2 μ (3.1738) u v = ( 2 μ r ) e ( r 2 μ ) / 2 μ {:(3.1738)uv=(2mu-r)e^((r-2mu)//2mu):}\begin{equation*} u v=(2 \mu-r) e^{(r-2 \mu) / 2 \mu} \tag{3.1738} \end{equation*}(3.1738)uv=(2μr)e(r2μ)/2μ
Note that f ( x ) = x e x / a f ( x ) = x e x / a f(x)=xe^(x//a)f(x)=x e^{x / a}f(x)=xex/a is monotonically increasing when x > a x > a x > -ax>-ax>a (and f ( x ) > f ( x ) > f(x) >f(x)>f(x)> a / e ) a / e ) -a//e)-a / e)a/e), and therefore, y = f ( x ) y = f ( x ) y=f(x)y=f(x)y=f(x) has a unique solution x x xxx for any y > a / e y > a / e y > -a//ey>-a / ey>a/e. We treat u u uuu as a kind of universal time and a timelike vector is future directed if its
Figure 3.16 The four regions K 1 , K 2 , K 3 K 1 , K 2 , K 3 K_(1),K_(2),K_(3)K_{1}, K_{2}, K_{3}K1,K2,K3, and K 4 K 4 K_(4)K_{4}K4 of Kruskal-Szekeres coordinates u u uuu and v v vvv.
projection to u u del_(u)\partial_{u}u is positive. The orientation (needed in integration) is defined by the ordering ( u , v , θ , ϕ u , v , θ , ϕ u,v,theta,phiu, v, \theta, \phiu,v,θ,ϕ ) of coordinates. Note that the radial null lines (radial light rays) are given by d u = 0 d u = 0 du=0d u=0du=0 or d v = 0 d v = 0 dv=0d v=0dv=0.
The Kruskal-Szekeres spacetime can be divided into four regions (see Figure 3.16): region K 1 K 1 K_(1)K_{1}K1 consists of points u < 0 , v > 0 u < 0 , v > 0 u < 0,v > 0u<0, v>0u<0,v>0, region K 2 K 2 K_(2)K_{2}K2 of points u , v > 0 u , v > 0 u,v > 0u, v>0u,v>0, region K 3 K 3 K_(3)K_{3}K3 of points u , v < 0 u , v < 0 u,v < 0u, v<0u,v<0, and finally, region K 4 K 4 K_(4)K_{4}K4 of points u > 0 , v < 0 u > 0 , v < 0 u > 0,v < 0u>0, v<0u>0,v<0. The boundaries between these regions are nonsingular points for the metric. The only singularities are at the boundary u v = 2 μ / e u v = 2 μ / e uv=2mu//eu v=2 \mu / euv=2μ/e. The region K 1 K 1 K_(1)K_{1}K1 is equivalent to the outer region of a Schwarzschild spacetime. This is seen by performing the coordinate transformation ( u , v , θ , ϕ ) ( t , r , θ , ϕ ) ( u , v , θ , ϕ ) ( t , r , θ , ϕ ) (u,v,theta,phi)|->(t,r,theta,phi)(u, v, \theta, \phi) \mapsto(t, r, \theta, \phi)(u,v,θ,ϕ)(t,r,θ,ϕ), where r = r ( u , v ) r = r ( u , v ) r=r(u,v)r=r(u, v)r=r(u,v) as above and the Schwarzschild time is t = 2 μ ln ( v / u ) t = 2 μ ln ( v / u ) t=2mu ln(-v//u)t=2 \mu \ln (-v / u)t=2μln(v/u). With a similar coordinate transformation, the region K 4 K 4 K_(4)K_{4}K4 is also seen to be equivalent to the outer Schwarzschild solution. The region K 2 K 2 K_(2)K_{2}K2 is equivalent with the Schwarzschild black hole. This equivalence is obtained through the coordinate transformation ( u , v , θ , ϕ ) ( t , r , θ , ϕ ) ( u , v , θ , ϕ ) ( t , r , θ , ϕ ) (u,v,theta,phi)|->(t,r,theta,phi)(u, v, \theta, \phi) \mapsto(t, r, \theta, \phi)(u,v,θ,ϕ)(t,r,θ,ϕ), where r = r ( u , v ) r = r ( u , v ) r=r(u,v)r=r(u, v)r=r(u,v) is the same as before but now t = 2 μ ln ( v / u ) t = 2 μ ln ( v / u ) t=2mu ln(v//u)t=2 \mu \ln (v / u)t=2μln(v/u). The region K 3 K 3 K_(3)K_{3}K3 is called a "white hole."
It is easy to construct smooth timelike curves, which go from either K 1 K 1 K_(1)K_{1}K1 or K 4 K 4 K_(4)K_{4}K4 to the black hole K 2 K 2 K_(2)K_{2}K2. However, we will prove that once an observer falls into the black hole K 2 K 2 K_(2)K_{2}K2, there is no way to go back to the "normal" regions K 1 K 1 K_(1)K_{1}K1 and K 4 K 4 K_(4)K_{4}K4. Analogously, everything escapes the "white hole" K 3 K 3 K_(3)K_{3}K3.
Let x ( t ) x ( t ) x(t)x(t)x(t) be the timelike path of an observer. Then, along the path
(3.1739) d r d t = r u d u d t + r v d v d t = r 8 μ 2 e ( r 2 μ ) / 2 μ [ r u g ( v , x ( t ) ) + r v g ( u , x ( t ) ) ] < 0 (3.1739) d r d t = r u d u d t + r v d v d t = r 8 μ 2 e ( r 2 μ ) / 2 μ r u g v , x ( t ) + r v g u , x ( t ) < 0 {:(3.1739)(dr)/(dt)=(del r)/(del u)*(du)/(dt)+(del r)/(del v)*(dv)/(dt)=(r)/(8mu^(2))e^((r-2mu)//2mu)[(del r)/(del u)g(del_(v),x^(')(t))+(del r)/(del v)g(del_(u),x^(')(t))] < 0:}\begin{equation*} \frac{d r}{d t}=\frac{\partial r}{\partial u} \cdot \frac{d u}{d t}+\frac{\partial r}{\partial v} \cdot \frac{d v}{d t}=\frac{r}{8 \mu^{2}} e^{(r-2 \mu) / 2 \mu}\left[\frac{\partial r}{\partial u} g\left(\partial_{v}, x^{\prime}(t)\right)+\frac{\partial r}{\partial v} g\left(\partial_{u}, x^{\prime}(t)\right)\right]<0 \tag{3.1739} \end{equation*}(3.1739)drdt=rududt+rvdvdt=r8μ2e(r2μ)/2μ[rug(v,x(t))+rvg(u,x(t))]<0
since x ( t ) x ( t ) x(t)x(t)x(t) is timelike and in K 2 K 2 K_(2)K_{2}K2 it holds that r r u = 2 μ v e ( 2 μ r ) / 2 μ < 0 r r u = 2 μ v e ( 2 μ r ) / 2 μ < 0 r(del r)/(del u)=-2mu ve^((2mu-r)//2mu) < 0r \frac{\partial r}{\partial u}=-2 \mu v e^{(2 \mu-r) / 2 \mu}<0rru=2μve(2μr)/2μ<0 and similarly for the v v vvv-coordinate.
The boundary between K 2 K 2 K_(2)K_{2}K2 and the normal regions is r = 2 μ r = 2 μ r=2mur=2 \mur=2μ (i.e., u = 0 u = 0 u=0u=0u=0 or v = 0 v = 0 v=0v=0v=0 ). The function r ( x ( t ) ) r ( x ( t ) ) r(x(t))r(x(t))r(x(t)) was seen to be decreasing, and therefore, the path x ( t ) x ( t ) x(t)x(t)x(t) can never hit the boundary r = 2 μ r = 2 μ r=2mur=2 \mur=2μ. However, the observer entering K 2 K 2 K_(2)K_{2}K2 has a deplorable future, since it will eventually hit the true singularity r = 0 r = 0 r=0r=0r=0, again using the monotonicity of the function r ( x ( t ) ) r ( x ( t ) ) r(x(t))r(x(t))r(x(t)).
Note that there is also another singularity, the outer boundary of K 3 K 3 K_(3)K_{3}K3. Nevertheless, this is of no great concern, since it is in the past; no future directed timelike curve can enter that singularity.

2.109

For s = 0 s = 0 s=0s=0s=0, we have r = 2 μ , v = v 0 , u = 0 r = 2 μ , v = v 0 , u = 0 r=2mu,v=v_(0),u=0r=2 \mu, v=v_{0}, u=0r=2μ,v=v0,u=0 and the constant of motion is then given by
(3.1740) u ˙ ( 0 ) v 0 2 μ = E 2 μ (3.1740) u ˙ ( 0 ) v 0 2 μ = E 2 μ {:(3.1740)((u^(˙))(0)v_(0))/(2mu)=(E)/(2mu):}\begin{equation*} \frac{\dot{u}(0) v_{0}}{2 \mu}=\frac{E}{2 \mu} \tag{3.1740} \end{equation*}(3.1740)u˙(0)v02μ=E2μ
It follows that, at any point,
(3.1741) E = 2 μ r e 2 μ r 2 μ ( u ˙ v v ˙ u ) (3.1741) E = 2 μ r e 2 μ r 2 μ ( u ˙ v v ˙ u ) {:(3.1741)E=(2mu)/(r)*e^((2mu-r)/(2mu))(u^(˙)v-v^(˙)u):}\begin{equation*} E=\frac{2 \mu}{r} \cdot e^{\frac{2 \mu-r}{2 \mu}}(\dot{u} v-\dot{v} u) \tag{3.1741} \end{equation*}(3.1741)E=2μre2μr2μ(u˙vv˙u)
Taking r r rrr as a new parameter of the geodesic and differentiating u v u v uvu vuv with respect to r r rrr, we obtain
(3.1742) d ( u v ) d r = d s d r ( u ˙ v + v ˙ u ) = d d r ( 2 μ r ) e r 2 μ 2 μ = r 2 μ e r 2 μ 2 μ . (3.1742) d ( u v ) d r = d s d r ( u ˙ v + v ˙ u ) = d d r ( 2 μ r ) e r 2 μ 2 μ = r 2 μ e r 2 μ 2 μ . {:(3.1742)(d(uv))/(dr)=(ds)/(dr)(u^(˙)v+v^(˙)u)=(d)/(dr)(2mu-r)e^((r-2mu)/(2mu))=-(r)/(2mu)e^((r-2mu)/(2mu)).:}\begin{equation*} \frac{d(u v)}{d r}=\frac{d s}{d r}(\dot{u} v+\dot{v} u)=\frac{d}{d r}(2 \mu-r) e^{\frac{r-2 \mu}{2 \mu}}=-\frac{r}{2 \mu} e^{\frac{r-2 \mu}{2 \mu}} . \tag{3.1742} \end{equation*}(3.1742)d(uv)dr=dsdr(u˙v+v˙u)=ddr(2μr)er2μ2μ=r2μer2μ2μ.
We now have the linear system
(3.1743) u ˙ v v ˙ u = E r 2 μ e r 2 μ 2 μ , (3.1744) u ˙ v + v ˙ u = 1 ( d s d r ) r 2 μ e r 2 μ 2 μ , (3.1743) u ˙ v v ˙ u = E r 2 μ e r 2 μ 2 μ , (3.1744) u ˙ v + v ˙ u = 1 d s d r r 2 μ e r 2 μ 2 μ , {:[(3.1743)u^(˙)v-v^(˙)u=E(r)/(2mu)e^((r-2mu)/(2mu))","],[(3.1744)u^(˙)v+v^(˙)u=-(1)/(((ds)/(dr)))(r)/(2mu)e^((r-2mu)/(2mu))","]:}\begin{align*} & \dot{u} v-\dot{v} u=E \frac{r}{2 \mu} e^{\frac{r-2 \mu}{2 \mu}}, \tag{3.1743}\\ & \dot{u} v+\dot{v} u=-\frac{1}{\left(\frac{d s}{d r}\right)} \frac{r}{2 \mu} e^{\frac{r-2 \mu}{2 \mu}}, \tag{3.1744} \end{align*}(3.1743)u˙vv˙u=Er2μer2μ2μ,(3.1744)u˙v+v˙u=1(dsdr)r2μer2μ2μ,
for u ˙ v u ˙ v u^(˙)v\dot{u} vu˙v and v ˙ u v ˙ u v^(˙)u\dot{v} uv˙u. Solving this system, we obtain
(3.1745) u ˙ v = r 4 μ e r 2 μ 2 μ [ E 1 ( d s d r ) ] , v ˙ u = r 4 μ e r 2 μ 2 μ [ E + 1 ( d s d r ) ] . (3.1745) u ˙ v = r 4 μ e r 2 μ 2 μ E 1 d s d r , v ˙ u = r 4 μ e r 2 μ 2 μ E + 1 d s d r . {:(3.1745)u^(˙)v=(r)/(4mu)e^((r-2mu)/(2mu))[E-(1)/(((ds)/(dr)))]","quadv^(˙)u=-(r)/(4mu)e^((r-2mu)/(2mu))[E+(1)/(((ds)/(dr)))].:}\begin{equation*} \dot{u} v=\frac{r}{4 \mu} e^{\frac{r-2 \mu}{2 \mu}}\left[E-\frac{1}{\left(\frac{d s}{d r}\right)}\right], \quad \dot{v} u=-\frac{r}{4 \mu} e^{\frac{r-2 \mu}{2 \mu}}\left[E+\frac{1}{\left(\frac{d s}{d r}\right)}\right] . \tag{3.1745} \end{equation*}(3.1745)u˙v=r4μer2μ2μ[E1(dsdr)],v˙u=r4μer2μ2μ[E+1(dsdr)].
In addition, we also have the requirement that
(3.1746) g i j x ˙ i x ˙ j = 1 16 μ 2 r e 2 μ r 2 μ u ˙ v v ˙ u u v = 1 (3.1746) g i j x ˙ i x ˙ j = 1 16 μ 2 r e 2 μ r 2 μ u ˙ v v ˙ u u v = 1 {:(3.1746)g_(ij)x^(˙)^(i)x^(˙)^(j)=1quad=>quad(16mu^(2))/(r)e^((2mu-r)/(2mu))((u^(˙))v(v^(˙))u)/(uv)=1:}\begin{equation*} g_{i j} \dot{x}^{i} \dot{x}^{j}=1 \quad \Rightarrow \quad \frac{16 \mu^{2}}{r} e^{\frac{2 \mu-r}{2 \mu}} \frac{\dot{u} v \dot{v} u}{u v}=1 \tag{3.1746} \end{equation*}(3.1746)gijx˙ix˙j=116μ2re2μr2μu˙vv˙uuv=1
Inserting our expressions for u ˙ v , v ˙ u u ˙ v , v ˙ u u^(˙)v,v^(˙)u\dot{u} v, \dot{v} uu˙v,v˙u, and the relation between u v u v uvu vuv and r r rrr, this is equivalent to
(3.1747) r 2 μ r [ 1 ( d s d r ) 2 E 2 ] = 1 (3.1747) r 2 μ r 1 d s d r 2 E 2 = 1 {:(3.1747)(r)/(2mu-r)[(1)/(((ds)/(dr))^(2))-E^(2)]=1:}\begin{equation*} \frac{r}{2 \mu-r}\left[\frac{1}{\left(\frac{d s}{d r}\right)^{2}}-E^{2}\right]=1 \tag{3.1747} \end{equation*}(3.1747)r2μr[1(dsdr)2E2]=1
Solving for d s / d r d s / d r ds//drd s / d rds/dr, we obtain
(3.1748) d s d r = 1 E 2 1 + 2 μ r (3.1748) d s d r = 1 E 2 1 + 2 μ r {:(3.1748)(ds)/(dr)=-(1)/(sqrt(E^(2)-1+(2mu)/(r))):}\begin{equation*} \frac{d s}{d r}=-\frac{1}{\sqrt{E^{2}-1+\frac{2 \mu}{r}}} \tag{3.1748} \end{equation*}(3.1748)dsdr=1E21+2μr
where we have used the negative root since the r r rrr-coordinate decreases when the proper time increases.
It follows that the proper time Δ s Δ s Delta s\Delta sΔs to reach the singularity is given by
(3.1749) Δ s = 0 Δ s d s = 2 μ 0 d s d r d r = 0 2 μ d r E 2 1 + 2 μ r (3.1749) Δ s = 0 Δ s d s = 2 μ 0 d s d r d r = 0 2 μ d r E 2 1 + 2 μ r {:(3.1749)Delta s=int_(0)^(Delta s)ds=int_(2mu)^(0)(ds)/(dr)dr=int_(0)^(2mu)(dr)/(sqrt(E^(2)-1+(2mu)/(r))):}\begin{equation*} \Delta s=\int_{0}^{\Delta s} d s=\int_{2 \mu}^{0} \frac{d s}{d r} d r=\int_{0}^{2 \mu} \frac{d r}{\sqrt{E^{2}-1+\frac{2 \mu}{r}}} \tag{3.1749} \end{equation*}(3.1749)Δs=0Δsds=2μ0dsdrdr=02μdrE21+2μr
and thus, we find that
(3.1750) f ( r ) = 1 E 2 1 + 2 μ r (3.1750) f ( r ) = 1 E 2 1 + 2 μ r {:(3.1750)f(r)=(1)/(sqrt(E^(2)-1+(2mu)/(r))):}\begin{equation*} f(r)=\frac{1}{\sqrt{E^{2}-1+\frac{2 \mu}{r}}} \tag{3.1750} \end{equation*}(3.1750)f(r)=1E21+2μr

2.110

a) The scalar quantity R μ ν α β R μ v α β R μ ν α β R μ v α β R_(mu nu alpha beta)R^(mu v alpha beta)R_{\mu \nu \alpha \beta} R^{\mu v \alpha \beta}RμναβRμvαβ has a singularity at r = 0 r = 0 r=0r=0r=0. From this follows that the singularity at r = 0 r = 0 r=0r=0r=0 is a physical singularity and not simply due to a bad choice of coordinates (which is the case for the coordinate singularity at r = r r = r r=r_(**)r=r_{*}r=r ).
b) For a radial light signal, we have d θ = d ϕ = 0 d θ = d ϕ = 0 d theta=d phi=0d \theta=d \phi=0dθ=dϕ=0 and d s 2 = 0 d s 2 = 0 ds^(2)=0d s^{2}=0ds2=0. It follows that
(3.1751) d r = ± ( 1 r r ) d t (3.1751) d r = ± 1 r r d t {:(3.1751)dr=+-(1-(r_(**))/(r))dt:}\begin{equation*} d r= \pm\left(1-\frac{r_{*}}{r}\right) d t \tag{3.1751} \end{equation*}(3.1751)dr=±(1rr)dt
For r > r r > r r > r_(**)r>r_{*}r>r, we can see from the metric that t t ttt is the time-coordinate since g t t > 0 g t t > 0 g_(tt) > 0g_{t t}>0gtt>0. The forward light cone is therefore given by
(3.1752) d r d t = ± ( 1 r r ) (3.1752) d r d t = ± 1 r r {:(3.1752)(dr)/(dt)=+-(1-(r_(**))/(r)):}\begin{equation*} \frac{d r}{d t}= \pm\left(1-\frac{r_{*}}{r}\right) \tag{3.1752} \end{equation*}(3.1752)drdt=±(1rr)
i.e., one side of the cone going radially outward and another radially inward. For r < r r < r r < r_(**)r<r_{*}r<r, we instead have g r r > 0 g r r > 0 g_(rr) > 0g_{r r}>0grr>0, and thus, r r rrr represents the time-coordinate with time increasing in the negative r r rrr-direction. Thus, for this case, we have
(3.1753) d t d r = ± ( 1 r r ) 1 (3.1753) d t d r = ± 1 r r 1 {:(3.1753)(dt)/(dr)=+-(1-(r_(**))/(r))^(-1):}\begin{equation*} \frac{d t}{d r}= \pm\left(1-\frac{r_{*}}{r}\right)^{-1} \tag{3.1753} \end{equation*}(3.1753)dtdr=±(1rr)1
i.e., the t t ttt-coordinate (which is now a spatial coordinate) can both decrease or increase with time ( r ) ( r ) (r)(r)(r).
c) In Kruskal-Szekeres coordinates, we still have d Ω = 0 d Ω = 0 d Omega=0d \Omega=0dΩ=0 and d s 2 = 0 d s 2 = 0 ds^(2)=0d s^{2}=0ds2=0 for the radial light cone. It follows that
(3.1754) d U = ± d V (3.1754) d U = ± d V {:(3.1754)dU=+-dV:}\begin{equation*} d U= \pm d V \tag{3.1754} \end{equation*}(3.1754)dU=±dV
Hence, the light cones are straight lines with slope 1 in these coordinates.

2.111

a) The equation of motion for a massive test particle m m mmm (at position r r r\mathbf{r}r ) in a gravitational potential Φ ( r ) = G M / r Φ ( r ) = G M / r Phi(r)=-GM//r\Phi(r)=-G M / rΦ(r)=GM/r according to Newton's mechanics (i.e., Newton's second law) is given by
(3.1755) F ( r ) = m r ¨ = m Φ ( r ) = m G M 1 r = m G M r 2 e r , (3.1755) F ( r ) = m r ¨ = m Φ ( r ) = m G M 1 r = m G M r 2 e r , {:(3.1755)F(r)=mr^(¨)=-m grad Phi(r)=mGM grad(1)/(r)=-(mGM)/(r^(2))e_(r)",":}\begin{equation*} \mathbf{F}(\mathbf{r})=m \ddot{\mathbf{r}}=-m \nabla \Phi(r)=m G M \nabla \frac{1}{r}=-\frac{m G M}{r^{2}} \mathbf{e}_{r}, \tag{3.1755} \end{equation*}(3.1755)F(r)=mr¨=mΦ(r)=mGM1r=mGMr2er,
where M M MMM is the total point mass of a spherically symmetric source (located at the origin) giving rise to the gravitational potential Φ ( r ) Φ ( r ) Phi(r)\Phi(r)Φ(r) (solving Newton's field equation in differential form, i.e., 2 Φ = 4 π G ρ 2 Φ = 4 π G ρ grad^(2)Phi=4pi G rho\nabla^{2} \Phi=4 \pi G \rho2Φ=4πGρ with ρ ρ rho\rhoρ being the mass density function), or without the massive test particle, the equation of motion in differential form is
(3.1756) d 2 r d t 2 = Φ ( r ) (3.1756) d 2 r d t 2 = Φ ( r ) {:(3.1756)(d^(2)r)/(dt^(2))=-grad Phi(r):}\begin{equation*} \frac{d^{2} \mathbf{r}}{d t^{2}}=-\nabla \Phi(r) \tag{3.1756} \end{equation*}(3.1756)d2rdt2=Φ(r)
whereas the equations of motion according to general relativity are given by the geodesic equations x ¨ μ + Γ ν λ μ x ˙ v x ˙ λ = 0 x ¨ μ + Γ ν λ μ x ˙ v x ˙ λ = 0 x^(¨)^(mu)+Gamma_(nu lambda)^(mu)x^(˙)^(v)x^(˙)^(lambda)=0\ddot{x}^{\mu}+\Gamma_{\nu \lambda}^{\mu} \dot{x}^{v} \dot{x}^{\lambda}=0x¨μ+Γνλμx˙vx˙λ=0, or more explicitly,
(3.1757) d 2 x μ d σ 2 + Γ v λ μ d x v d σ d x λ d σ = 0 (3.1757) d 2 x μ d σ 2 + Γ v λ μ d x v d σ d x λ d σ = 0 {:(3.1757)(d^(2)x^(mu))/(dsigma^(2))+Gamma_(v lambda)^(mu)(dx^(v))/(d sigma)*(dx^(lambda))/(d sigma)=0:}\begin{equation*} \frac{d^{2} x^{\mu}}{d \sigma^{2}}+\Gamma_{v \lambda}^{\mu} \frac{d x^{v}}{d \sigma} \cdot \frac{d x^{\lambda}}{d \sigma}=0 \tag{3.1757} \end{equation*}(3.1757)d2xμdσ2+Γvλμdxvdσdxλdσ=0
where σ σ sigma\sigmaσ is the curve parameter and Γ ν λ μ Γ ν λ μ Gamma_(nu lambda)^(mu)\Gamma_{\nu \lambda}^{\mu}Γνλμ are the Christoffel symbols defined through
(3.1758) g α μ Γ ν λ μ = 1 2 ( g v α x λ + g λ α x v g ν λ x α ) (3.1758) g α μ Γ ν λ μ = 1 2 g v α x λ + g λ α x v g ν λ x α {:(3.1758)g_(alpha mu)Gamma_(nu lambda)^(mu)=(1)/(2)((delg_(v alpha))/(delx^(lambda))+(delg_(lambda alpha))/(delx^(v))-(delg_(nu lambda))/(delx^(alpha))):}\begin{equation*} g_{\alpha \mu} \Gamma_{\nu \lambda}^{\mu}=\frac{1}{2}\left(\frac{\partial g_{v \alpha}}{\partial x^{\lambda}}+\frac{\partial g_{\lambda \alpha}}{\partial x^{v}}-\frac{\partial g_{\nu \lambda}}{\partial x^{\alpha}}\right) \tag{3.1758} \end{equation*}(3.1758)gαμΓνλμ=12(gvαxλ+gλαxvgνλxα)
with g μ ν g μ ν g_(mu nu)g_{\mu \nu}gμν being the metric. Thus, the Christoffel symbols are combinations of firstorder derivatives of the metric.
In the Newtonian limit, we wish to derive the equation of motion according to Newton's mechanics from the equations of motion according to general relativity. Thus, we compute the geodesic equations using the metric g μ ν = η μ ν + g μ ν = η μ ν + g_(mu nu)=eta_(mu nu)+g_{\mu \nu}=\eta_{\mu \nu}+gμν=ημν+ h μ ν h μ ν h_(mu nu)h_{\mu \nu}hμν in the linear approximation, i.e., we neglect higher-order terms in h μ ν h μ ν h_(mu nu)h_{\mu \nu}hμν. Note that ( η μ ν ) = diag ( 1 , 1 , 1 , 1 ) η μ ν = diag ( 1 , 1 , 1 , 1 ) (eta_(mu nu))=diag(1,-1,-1,-1)\left(\eta_{\mu \nu}\right)=\operatorname{diag}(1,-1,-1,-1)(ημν)=diag(1,1,1,1). For small velocities, the time component x ˙ 0 ( σ ) x ˙ 0 ( σ ) x^(˙)^(0)(sigma)\dot{x}^{0}(\sigma)x˙0(σ) of the 4 -velocity is much larger than the spatial components. For this reason, we can approximate the geodesic equations as
(3.1759) d 2 x μ d σ 2 + Γ 00 μ ( d x 0 d σ ) 2 = 0 (3.1759) d 2 x μ d σ 2 + Γ 00 μ d x 0 d σ 2 = 0 {:(3.1759)(d^(2)x^(mu))/(dsigma^(2))+Gamma_(00)^(mu)((dx^(0))/(d sigma))^(2)=0:}\begin{equation*} \frac{d^{2} x^{\mu}}{d \sigma^{2}}+\Gamma_{00}^{\mu}\left(\frac{d x^{0}}{d \sigma}\right)^{2}=0 \tag{3.1759} \end{equation*}(3.1759)d2xμdσ2+Γ00μ(dx0dσ)2=0
In the linear approximation, we have
(3.1760) Γ 00 0 = 1 c 2 0 Φ , Γ 00 i = 1 c 2 i Φ (3.1760) Γ 00 0 = 1 c 2 0 Φ , Γ 00 i = 1 c 2 i Φ {:(3.1760)Gamma_(00)^(0)=(1)/(c^(2))del^(0)Phi","quadGamma_(00)^(i)=-(1)/(c^(2))del^(i)Phi:}\begin{equation*} \Gamma_{00}^{0}=\frac{1}{c^{2}} \partial^{0} \Phi, \quad \Gamma_{00}^{i}=-\frac{1}{c^{2}} \partial^{i} \Phi \tag{3.1760} \end{equation*}(3.1760)Γ000=1c20Φ,Γ00i=1c2iΦ
Thus, the geodesic equations become
(3.1761) x ¨ 0 + 1 c 2 0 Φ ( x ˙ 0 ) 2 = 0 (3.1762) x ¨ i 1 c 2 i Φ ( x ˙ 0 ) 2 = 0 (3.1761) x ¨ 0 + 1 c 2 0 Φ x ˙ 0 2 = 0 (3.1762) x ¨ i 1 c 2 i Φ x ˙ 0 2 = 0 {:[(3.1761)x^(¨)^(0)+(1)/(c^(2))*del^(0)Phi(x^(˙)^(0))^(2)=0],[(3.1762)x^(¨)^(i)-(1)/(c^(2))*del^(i)Phi(x^(˙)^(0))^(2)=0]:}\begin{align*} & \ddot{x}^{0}+\frac{1}{c^{2}} \cdot \partial^{0} \Phi\left(\dot{x}^{0}\right)^{2}=0 \tag{3.1761}\\ & \ddot{x}^{i}-\frac{1}{c^{2}} \cdot \partial^{i} \Phi\left(\dot{x}^{0}\right)^{2}=0 \tag{3.1762} \end{align*}(3.1761)x¨0+1c20Φ(x˙0)2=0(3.1762)x¨i1c2iΦ(x˙0)2=0
In the frame, where the source is at rest, the first equation says that we can choose the time t t ttt as the curve parameter, i.e., x 0 ( σ ) = σ = c t x 0 ( σ ) = σ = c t x^(0)(sigma)=sigma=ctx^{0}(\sigma)=\sigma=c tx0(σ)=σ=ct, and then the second equation becomes
(3.1763) d 2 x i d t 2 = i Φ = i Φ d 2 x d t 2 = Φ (3.1763) d 2 x i d t 2 = i Φ = i Φ d 2 x d t 2 = Φ {:(3.1763)(d^(2)x^(i))/(dt^(2))=del^(i)Phi=-del_(i)Phiquad<=>quad(d^(2)x)/(dt^(2))=-grad Phi:}\begin{equation*} \frac{d^{2} x^{i}}{d t^{2}}=\partial^{i} \Phi=-\partial_{i} \Phi \quad \Leftrightarrow \quad \frac{d^{2} \mathbf{x}}{d t^{2}}=-\nabla \Phi \tag{3.1763} \end{equation*}(3.1763)d2xidt2=iΦ=iΦd2xdt2=Φ
The right-hand side (after multiplication by the mass m m mmm of the test particle) is the gravitational force of the source on m m mmm, so this equation is just Newton's second law, i.e., m r ¨ = m Φ ( r ) m r ¨ = m Φ ( r ) mr^(¨)=-m grad Phi(r)m \ddot{\mathbf{r}}=-m \nabla \Phi(r)mr¨=mΦ(r).
b) Tidal forces in Newtonian physics are due to differences in gravitational accelerations in neighboring points. Since gravitational accelerations are proportional to the derivatives of the gravitational potential, the differences in gravitational accelerations are second-order derivatives of the gravitational potential, and therefore proportional to r 3 r 3 r^(-3)r^{-3}r3, since the gravitational potential itself is proportional to r 1 r 1 r^(-1)r^{-1}r1. In general relativity, tidal forces are related to the so-called geodesic deviation, which describes how nearby geodesics separate or converge. They are proportional to second-order derivatives of the metric, since the relativistic gravitational potential in the weak field limit appears as the perturbations to the metric. Thus, tidal forces in general relativity are proportional to the curvature (which is proportional to second-order derivatives of the metric).

2.112

a) In the Newtonian limit (where | h μ ν | 1 h μ ν 1 |h_(mu nu)|≪1\left|h_{\mu \nu}\right| \ll 1|hμν|1 everywhere in spacetime), we have the metric tensor g μ ν g μ ν g_(mu nu)g_{\mu \nu}gμν, the Christoffel symbols Γ μ ν λ Γ μ ν λ Gamma_(mu nu)^(lambda)\Gamma_{\mu \nu}^{\lambda}Γμνλ, the Riemann curvature tensor R μ ν λ ρ R μ ν λ ρ R^(mu)_(nu lambda rho)R^{\mu}{ }_{\nu \lambda \rho}Rμνλρ, the Ricci tensor R μ ν R μ ν R_(mu nu)R_{\mu \nu}Rμν, the Ricci scalar R R RRR, and the Einstein tensor G μ ν G μ ν G_(mu nu)G_{\mu \nu}Gμν as
(3.1764) g μ ν = η μ ν + h μ ν Γ μ ν λ = 1 2 g λ ρ ( μ g ν ρ + ν g μ ρ ρ g μ ν ) (3.1765) 1 2 η λ ρ ( μ h ν ρ + ν h μ ρ ρ h μ ν ) Γ μ ν ( 1 ) μ ν λ (3.1764) g μ ν = η μ ν + h μ ν Γ μ ν λ = 1 2 g λ ρ μ g ν ρ + ν g μ ρ ρ g μ ν (3.1765) 1 2 η λ ρ μ h ν ρ + ν h μ ρ ρ h μ ν Γ μ ν ( 1 ) μ ν λ {:[(3.1764)g_(mu nu)=eta_(mu nu)+h_(mu nu)],[Gamma_(mu nu)^(lambda)=(1)/(2)g^(lambda rho)(del_(mu)g_(nu rho)+del_(nu)g_(mu rho)-del_(rho)g_(mu nu))],[(3.1765)≃(1)/(2)eta^(lambda rho)(del_(mu)h_(nu rho)+del_(nu)h_(mu rho)-del_(rho)h_(mu nu))-=Gamma_(mu nu)^((1))_(mu nu)^(lambda)]:}\begin{align*} g_{\mu \nu} & =\eta_{\mu \nu}+h_{\mu \nu} \tag{3.1764}\\ \Gamma_{\mu \nu}^{\lambda} & =\frac{1}{2} g^{\lambda \rho}\left(\partial_{\mu} g_{\nu \rho}+\partial_{\nu} g_{\mu \rho}-\partial_{\rho} g_{\mu \nu}\right) \\ & \simeq \frac{1}{2} \eta^{\lambda \rho}\left(\partial_{\mu} h_{\nu \rho}+\partial_{\nu} h_{\mu \rho}-\partial_{\rho} h_{\mu \nu}\right) \equiv \Gamma_{\mu \nu}^{(1)}{ }_{\mu \nu}^{\lambda} \tag{3.1765} \end{align*}(3.1764)gμν=ημν+hμνΓμνλ=12gλρ(μgνρ+νgμρρgμν)(3.1765)12ηλρ(μhνρ+νhμρρhμν)Γμν(1)μνλ
R μ ν λ ρ = λ Γ ρ ν μ ρ Γ λ ν μ + Γ λ σ μ Γ ρ ν σ Γ ρ σ μ λ ν σ λ Γ ( 1 ) ρ ν μ ρ Γ ( 1 ) μ λ ν = 1 2 η μ σ ( λ ρ h ν σ + ν λ h ρ σ λ σ h ρ v λ ρ h ν σ ν ρ h λ σ + ρ σ h λ ν ) = 1 2 ( ν λ h ρ μ + μ ρ h λ ν ν ρ h λ μ μ λ h ρ v ) (3.1766) = 1 2 η μ σ ( ν λ h ρ σ + σ ρ h ν λ ν ρ h λ σ σ λ h ν ρ ) R ( 1 ) μ ν λ , R μ ν λ ρ = g μ σ R σ ν λ ρ η μ σ R ( 1 ) σ ν λ ρ (3.1767) = 1 2 ( ν λ h μ ρ + μ ρ h ν λ ν ρ h μ λ μ λ h ν ρ ) R μ ν λ ρ ( 1 ) , R μ ν = g λ ρ R ρ μ λ ν = R λ μ λ ν η λ ρ R ρ μ λ ν ( 1 ) (3.1768) = 1 2 ( λ μ h v λ + λ ν h μ λ μ ν h h μ ν ) R μ ν ( 1 ) , (3.1769) R = g μ ν R μ ν = R μ μ η μ ν R μ ν ( 1 ) = μ ν h μ ν h R ( 1 ) , G μ ν = R μ ν 1 2 g μ ν R R μ ν ( 1 ) 1 2 η μ ν R ( 1 ) (3.1770) = 1 2 ( λ μ h v λ + λ ν h μ λ μ ν h h μ ν η μ ν α β h α β + η μ ν h ) G μ ν ( 1 ) , R μ ν λ ρ = λ Γ ρ ν μ ρ Γ λ ν μ + Γ λ σ μ Γ ρ ν σ Γ ρ σ μ λ ν σ λ Γ ( 1 ) ρ ν μ ρ Γ ( 1 ) μ λ ν = 1 2 η μ σ λ ρ h ν σ + ν λ h ρ σ λ σ h ρ v λ ρ h ν σ ν ρ h λ σ + ρ σ h λ ν = 1 2 ν λ h ρ μ + μ ρ h λ ν ν ρ h λ μ μ λ h ρ v (3.1766) = 1 2 η μ σ ν λ h ρ σ + σ ρ h ν λ ν ρ h λ σ σ λ h ν ρ R ( 1 ) μ ν λ , R μ ν λ ρ = g μ σ R σ ν λ ρ η μ σ R ( 1 ) σ ν λ ρ (3.1767) = 1 2 ν λ h μ ρ + μ ρ h ν λ ν ρ h μ λ μ λ h ν ρ R μ ν λ ρ ( 1 ) , R μ ν = g λ ρ R ρ μ λ ν = R λ μ λ ν η λ ρ R ρ μ λ ν ( 1 ) (3.1768) = 1 2 λ μ h v λ + λ ν h μ λ μ ν h h μ ν R μ ν ( 1 ) , (3.1769) R = g μ ν R μ ν = R μ μ η μ ν R μ ν ( 1 ) = μ ν h μ ν h R ( 1 ) , G μ ν = R μ ν 1 2 g μ ν R R μ ν ( 1 ) 1 2 η μ ν R ( 1 ) (3.1770) = 1 2 λ μ h v λ + λ ν h μ λ μ ν h h μ ν η μ ν α β h α β + η μ ν h G μ ν ( 1 ) , {:[R^(mu)_(nu lambda rho)=del_(lambda)Gamma_(rho nu)^(mu)-del_(rho)Gamma_(lambda nu)^(mu)+Gamma_(lambda sigma)^(mu)Gamma_(rho nu)^(sigma)-Gamma_(rho sigma)^(mu)_(lambda nu)^(sigma)≃del_(lambda)Gamma^((1))_(rho nu)^(mu)-del_(rho)Gamma^((1)mu)_(lambda nu)],[=(1)/(2)eta^(mu sigma)(del_(lambda)del_(rho)h_(nu sigma)+del_(nu)del_(lambda)h_(rho sigma)-del_(lambda)del_(sigma)h_(rho v)-del_(lambda)del_(rho)h_(nu sigma)-del_(nu)del_(rho)h_(lambda sigma)+del_(rho)del_(sigma)h_(lambda nu))],[=(1)/(2)(del_(nu)del_(lambda)h_(rho)^(mu)+del^(mu)del_(rho)h_(lambda nu)-del_(nu)del_(rho)h_(lambda)^(mu)-del^(mu)del_(lambda)h_(rho v))],[(3.1766)=(1)/(2)eta^(mu sigma)(del_(nu)del_(lambda)h_(rho sigma)+del_(sigma)del_(rho)h_(nu lambda)-del_(nu)del_(rho)h_(lambda sigma)-del_(sigma)del_(lambda)h_(nu rho))-=R^((1)^(mu)_(nu lambda),)],[R_(mu nu lambda rho)=g_(mu sigma)R^(sigma)_(nu lambda rho)≃eta_(mu sigma)R^((1)^(sigma))_(nu lambda rho)],[(3.1767)=(1)/(2)(del_(nu)del_(lambda)h_(mu rho)+del_(mu)del_(rho)h_(nu lambda)-del_(nu)del_(rho)h_(mu lambda)-del_(mu)del_(lambda)h_(nu rho))-=R_(mu nu lambda rho)^((1))","],[R_(mu nu)=g^(lambda rho)R_(rho mu lambda nu)=R^(lambda)_(mu lambda nu)≃eta^(lambda rho)R_(rho mu lambda nu)^((1))],[(3.1768)=(1)/(2)(del_(lambda)del_(mu)h_(v)^(lambda)+del_(lambda)del_(nu)h_(mu)^(lambda)-del_(mu)del_(nu)h-◻h_(mu nu))-=R_(mu nu)^((1))","],[(3.1769)R=g^(mu nu)R_(mu nu)=R_(mu)^(mu)≃eta^(mu nu)R_(mu nu)^((1))=del^(mu)del^(nu)h_(mu nu)-◻h-=R^((1))","],[G_(mu nu)=R_(mu nu)-(1)/(2)g_(mu nu)R≃R_(mu nu)^((1))-(1)/(2)eta_(mu nu)R^((1))],[(3.1770)=(1)/(2)(del_(lambda)del_(mu)h_(v)^(lambda)+del_(lambda)del_(nu)h_(mu)^(lambda)-del_(mu)del_(nu)h-◻h_(mu nu)-eta_(mu nu)del^(alpha)del^(beta)h_(alpha beta)+eta_(mu nu)◻h)-=G_(mu nu)^((1))","]:}\begin{align*} & R^{\mu}{ }_{\nu \lambda \rho}=\partial_{\lambda} \Gamma_{\rho \nu}^{\mu}-\partial_{\rho} \Gamma_{\lambda \nu}^{\mu}+\Gamma_{\lambda \sigma}^{\mu} \Gamma_{\rho \nu}^{\sigma}-\Gamma_{\rho \sigma}^{\mu}{ }_{\lambda \nu}^{\sigma} \simeq \partial_{\lambda} \Gamma^{(1)}{ }_{\rho \nu}^{\mu}-\partial_{\rho} \Gamma^{(1) \mu}{ }_{\lambda \nu} \\ & =\frac{1}{2} \eta^{\mu \sigma}\left(\partial_{\lambda} \partial_{\rho} h_{\nu \sigma}+\partial_{\nu} \partial_{\lambda} h_{\rho \sigma}-\partial_{\lambda} \partial_{\sigma} h_{\rho v}-\partial_{\lambda} \partial_{\rho} h_{\nu \sigma}-\partial_{\nu} \partial_{\rho} h_{\lambda \sigma}+\partial_{\rho} \partial_{\sigma} h_{\lambda \nu}\right) \\ & =\frac{1}{2}\left(\partial_{\nu} \partial_{\lambda} h_{\rho}^{\mu}+\partial^{\mu} \partial_{\rho} h_{\lambda \nu}-\partial_{\nu} \partial_{\rho} h_{\lambda}^{\mu}-\partial^{\mu} \partial_{\lambda} h_{\rho v}\right) \\ & =\frac{1}{2} \eta^{\mu \sigma}\left(\partial_{\nu} \partial_{\lambda} h_{\rho \sigma}+\partial_{\sigma} \partial_{\rho} h_{\nu \lambda}-\partial_{\nu} \partial_{\rho} h_{\lambda \sigma}-\partial_{\sigma} \partial_{\lambda} h_{\nu \rho}\right) \equiv R^{(1)^{\mu}{ }_{\nu \lambda},} \tag{3.1766}\\ & R_{\mu \nu \lambda \rho}=g_{\mu \sigma} R^{\sigma}{ }_{\nu \lambda \rho} \simeq \eta_{\mu \sigma} R^{(1)^{\sigma}}{ }_{\nu \lambda \rho} \\ & =\frac{1}{2}\left(\partial_{\nu} \partial_{\lambda} h_{\mu \rho}+\partial_{\mu} \partial_{\rho} h_{\nu \lambda}-\partial_{\nu} \partial_{\rho} h_{\mu \lambda}-\partial_{\mu} \partial_{\lambda} h_{\nu \rho}\right) \equiv R_{\mu \nu \lambda \rho}^{(1)}, \tag{3.1767}\\ & R_{\mu \nu}=g^{\lambda \rho} R_{\rho \mu \lambda \nu}=R^{\lambda}{ }_{\mu \lambda \nu} \simeq \eta^{\lambda \rho} R_{\rho \mu \lambda \nu}^{(1)} \\ & =\frac{1}{2}\left(\partial_{\lambda} \partial_{\mu} h_{v}^{\lambda}+\partial_{\lambda} \partial_{\nu} h_{\mu}^{\lambda}-\partial_{\mu} \partial_{\nu} h-\square h_{\mu \nu}\right) \equiv R_{\mu \nu}^{(1)}, \tag{3.1768}\\ & R=g^{\mu \nu} R_{\mu \nu}=R_{\mu}^{\mu} \simeq \eta^{\mu \nu} R_{\mu \nu}^{(1)}=\partial^{\mu} \partial^{\nu} h_{\mu \nu}-\square h \equiv R^{(1)}, \tag{3.1769}\\ & G_{\mu \nu}=R_{\mu \nu}-\frac{1}{2} g_{\mu \nu} R \simeq R_{\mu \nu}^{(1)}-\frac{1}{2} \eta_{\mu \nu} R^{(1)} \\ & =\frac{1}{2}\left(\partial_{\lambda} \partial_{\mu} h_{v}^{\lambda}+\partial_{\lambda} \partial_{\nu} h_{\mu}^{\lambda}-\partial_{\mu} \partial_{\nu} h-\square h_{\mu \nu}-\eta_{\mu \nu} \partial^{\alpha} \partial^{\beta} h_{\alpha \beta}+\eta_{\mu \nu} \square h\right) \equiv G_{\mu \nu}^{(1)}, \tag{3.1770} \end{align*}Rμνλρ=λΓρνμρΓλνμ+ΓλσμΓρνσΓρσμλνσλΓ(1)ρνμρΓ(1)μλν=12ημσ(λρhνσ+νλhρσλσhρvλρhνσνρhλσ+ρσhλν)=12(νλhρμ+μρhλννρhλμμλhρv)(3.1766)=12ημσ(νλhρσ+σρhνλνρhλσσλhνρ)R(1)μνλ,Rμνλρ=gμσRσνλρημσR(1)σνλρ(3.1767)=12(νλhμρ+μρhνλνρhμλμλhνρ)Rμνλρ(1),Rμν=gλρRρμλν=RλμλνηλρRρμλν(1)(3.1768)=12(λμhvλ+λνhμλμνhhμν)Rμν(1),(3.1769)R=gμνRμν=RμμημνRμν(1)=μνhμνhR(1),Gμν=Rμν12gμνRRμν(1)12ημνR(1)(3.1770)=12(λμhvλ+λνhμλμνhhμνημναβhαβ+ημνh)Gμν(1),
where h μ ν = h ν μ h μ ν = h ν μ h_(mu nu)=h_(nu mu)h_{\mu \nu}=h_{\nu \mu}hμν=hνμ and h h μ μ h h μ μ h-=h_(mu)^(mu)h \equiv h_{\mu}^{\mu}hhμμ. Note that in the weak field limit the Christoffel symbols are first-order derivatives of h μ ν h μ ν h_(mu nu)h_{\mu \nu}hμν, whereas the Riemann curvature tensor, the Ricci tensor, the Ricci scalar, and the Einstein tensor are all second-order derivatives of h μ ν h μ ν h_(mu nu)h_{\mu \nu}hμν.
Thus, the solution to this problem, i.e., the Ricci tensor in the linear approximation for a metric g μ ν = η μ ν + h μ ν g μ ν = η μ ν + h μ ν g_(mu nu)=eta_(mu nu)+h_(mu nu)g_{\mu \nu}=\eta_{\mu \nu}+h_{\mu \nu}gμν=ημν+hμν, is given by
(3.1771) R μ ν ( 1 ) = 1 2 ( λ μ h ν λ + λ ν h μ λ μ ν h h μ ν ) . (3.1771) R μ ν ( 1 ) = 1 2 λ μ h ν λ + λ ν h μ λ μ ν h h μ ν . {:(3.1771)R_(mu nu)^((1))=(1)/(2)(del_(lambda)del_(mu)h_(nu)^(lambda)+del_(lambda)del_(nu)h_(mu)^(lambda)-del_(mu)del_(nu)h-◻h_(mu nu)).:}\begin{equation*} R_{\mu \nu}^{(1)}=\frac{1}{2}\left(\partial_{\lambda} \partial_{\mu} h_{\nu}^{\lambda}+\partial_{\lambda} \partial_{\nu} h_{\mu}^{\lambda}-\partial_{\mu} \partial_{\nu} h-\square h_{\mu \nu}\right) . \tag{3.1771} \end{equation*}(3.1771)Rμν(1)=12(λμhνλ+λνhμλμνhhμν).
b) Consider the coordinate transformation x μ x μ = x μ + χ μ x μ x μ = x μ + χ μ x^(mu)|->x^('mu)=x^(mu)+chi^(mu)x^{\mu} \mapsto x^{\prime \mu}=x^{\mu}+\chi^{\mu}xμxμ=xμ+χμ, where | v χ μ | 1 v χ μ 1 |del_(v)chi^(mu)|≪1\left|\partial_{v} \chi^{\mu}\right| \ll 1|vχμ|1. Differentiating this coordinate transformation yields
(3.1772) ν x μ = ν x μ + ν χ μ = δ v μ + ν χ μ . (3.1772) ν x μ = ν x μ + ν χ μ = δ v μ + ν χ μ . {:(3.1772)del_(nu)x^('mu)=del_(nu)x^(mu)+del_(nu)chi^(mu)=delta_(v)^(mu)+del_(nu)chi^(mu).:}\begin{equation*} \partial_{\nu} x^{\prime \mu}=\partial_{\nu} x^{\mu}+\partial_{\nu} \chi^{\mu}=\delta_{v}^{\mu}+\partial_{\nu} \chi^{\mu} . \tag{3.1772} \end{equation*}(3.1772)νxμ=νxμ+νχμ=δvμ+νχμ.
Since | ν χ μ | 1 ν χ μ 1 |del_(nu)chi^(mu)|≪1\left|\partial_{\nu} \chi^{\mu}\right| \ll 1|νχμ|1, we have the inverse coordinate transformation, namely
(3.1773) v x μ δ v μ v χ μ . (3.1773) v x μ δ v μ v χ μ . {:(3.1773)del_(v)^(')x^(mu)≃delta_(v)^(mu)-del_(v)chi^(mu).:}\begin{equation*} \partial_{v}^{\prime} x^{\mu} \simeq \delta_{v}^{\mu}-\partial_{v} \chi^{\mu} . \tag{3.1773} \end{equation*}(3.1773)vxμδvμvχμ.
Using this inverse coordinate transformation for the metric tensor and the definition g μ ν ( 1 ) = η μ ν + h μ ν g μ ν ( 1 ) = η μ ν + h μ ν g_(mu nu)^((1))=eta_(mu nu)+h_(mu nu)g_{\mu \nu}^{(1)}=\eta_{\mu \nu}+h_{\mu \nu}gμν(1)=ημν+hμν, we find that
(3.1774) g μ ν = μ x α ν x β g α β δ μ α δ ν β g α β μ χ α η α ν ν χ β η μ β = g μ ν μ χ ν ν χ μ . (3.1774) g μ ν = μ x α ν x β g α β δ μ α δ ν β g α β μ χ α η α ν ν χ β η μ β = g μ ν μ χ ν ν χ μ . {:(3.1774)g_(mu nu)^(')=del_(mu)^(')x^(alpha)del_(nu)^(')x^(beta)g_(alpha beta)≃delta_(mu)^(alpha)delta_(nu)^(beta)g_(alpha beta)-del_(mu)chi^(alpha)eta_(alpha nu)-del_(nu)chi^(beta)eta_(mu beta)=g_(mu nu)-del_(mu)chi_(nu)-del_(nu)chi_(mu).:}\begin{equation*} g_{\mu \nu}^{\prime}=\partial_{\mu}^{\prime} x^{\alpha} \partial_{\nu}^{\prime} x^{\beta} g_{\alpha \beta} \simeq \delta_{\mu}^{\alpha} \delta_{\nu}^{\beta} g_{\alpha \beta}-\partial_{\mu} \chi^{\alpha} \eta_{\alpha \nu}-\partial_{\nu} \chi^{\beta} \eta_{\mu \beta}=g_{\mu \nu}-\partial_{\mu} \chi_{\nu}-\partial_{\nu} \chi_{\mu} . \tag{3.1774} \end{equation*}(3.1774)gμν=μxανxβgαβδμαδνβgαβμχαηαννχβημβ=gμνμχννχμ.
Finally, using again the definition g μ ν = η μ ν + h μ ν g μ ν = η μ ν + h μ ν g_(mu nu)=eta_(mu nu)+h_(mu nu)g_{\mu \nu}=\eta_{\mu \nu}+h_{\mu \nu}gμν=ημν+hμν, we obtain
(3.1775) h μ ν h μ ν = h μ ν μ χ ν ν χ μ (3.1775) h μ ν h μ ν = h μ ν μ χ ν ν χ μ {:(3.1775)h_(mu nu)|->h_(mu nu)^(')=h_(mu nu)-del_(mu)chi_(nu)-del_(nu)chi_(mu):}\begin{equation*} h_{\mu \nu} \mapsto h_{\mu \nu}^{\prime}=h_{\mu \nu}-\partial_{\mu} \chi_{\nu}-\partial_{\nu} \chi_{\mu} \tag{3.1775} \end{equation*}(3.1775)hμνhμν=hμνμχννχμ
which is the gauge transformation of h μ ν h μ ν h_(mu nu)h_{\mu \nu}hμν.
c) Consider a coordinate system in which the harmonic (or Lorenz) gauge condition holds, i.e., μ h ¯ μ ν = 0 μ h ¯ μ ν = 0 del^(mu) bar(h)_(mu nu)=0\partial^{\mu} \bar{h}_{\mu \nu}=0μh¯μν=0, where
(3.1776) h ¯ μ ν h μ ν 1 2 h η μ ν (3.1776) h ¯ μ ν h μ ν 1 2 h η μ ν {:(3.1776) bar(h)_(mu nu)-=h_(mu nu)-(1)/(2)heta_(mu nu):}\begin{equation*} \bar{h}_{\mu \nu} \equiv h_{\mu \nu}-\frac{1}{2} h \eta_{\mu \nu} \tag{3.1776} \end{equation*}(3.1776)h¯μνhμν12hημν
Note that h ¯ μ μ = h μ μ ( h ¯ μ μ = h μ μ ( bar(h)_(mu)^(mu)=-h_(mu)^(mu)(\bar{h}_{\mu}^{\mu}=-h_{\mu}^{\mu}(h¯μμ=hμμ( or h ¯ = h ) h ¯ = h ) bar(h)=-h)\bar{h}=-h)h¯=h). The tensor h ¯ μ ν h ¯ μ ν bar(h)_(mu nu)\bar{h}_{\mu \nu}h¯μν has the following gauge transformation
(3.1777) h ¯ μ v h ¯ μ ν = h ¯ μ ν μ χ ν ν χ μ + ( λ χ λ ) η μ v (3.1777) h ¯ μ v h ¯ μ ν = h ¯ μ ν μ χ ν ν χ μ + λ χ λ η μ v {:(3.1777) bar(h)_(mu v)|-> bar(h)_(mu nu)^(')= bar(h)_(mu nu)-del_(mu)chi_(nu)-del_(nu)chi_(mu)+(del_(lambda)chi^(lambda))eta_(mu v):}\begin{equation*} \bar{h}_{\mu v} \mapsto \bar{h}_{\mu \nu}^{\prime}=\bar{h}_{\mu \nu}-\partial_{\mu} \chi_{\nu}-\partial_{\nu} \chi_{\mu}+\left(\partial_{\lambda} \chi^{\lambda}\right) \eta_{\mu v} \tag{3.1777} \end{equation*}(3.1777)h¯μvh¯μν=h¯μνμχννχμ+(λχλ)ημv
Inserting the definition of h ¯ μ ν h ¯ μ ν bar(h)_(mu nu)\bar{h}_{\mu \nu}h¯μν into the harmonic gauge condition, we have
(3.1778) μ h μ ν = 1 2 ν h μ μ = 1 2 ν h (3.1778) μ h μ ν = 1 2 ν h μ μ = 1 2 ν h {:(3.1778)del^(mu)h_(mu nu)=(1)/(2)del_(nu)h_(mu)^(mu)=(1)/(2)del_(nu)h:}\begin{equation*} \partial^{\mu} h_{\mu \nu}=\frac{1}{2} \partial_{\nu} h_{\mu}^{\mu}=\frac{1}{2} \partial_{\nu} h \tag{3.1778} \end{equation*}(3.1778)μhμν=12νhμμ=12νh
which implies that
(3.1779) R μ ν ( 1 ) = 1 2 h μ ν , R ( 1 ) = 1 2 h , (3.1779) R μ ν ( 1 ) = 1 2 h μ ν , R ( 1 ) = 1 2 h , {:(3.1779)R_(mu nu)^((1))=-(1)/(2)◻h_(mu nu)","quadR^((1))=-(1)/(2)◻h",":}\begin{equation*} R_{\mu \nu}^{(1)}=-\frac{1}{2} \square h_{\mu \nu}, \quad R^{(1)}=-\frac{1}{2} \square h, \tag{3.1779} \end{equation*}(3.1779)Rμν(1)=12hμν,R(1)=12h,
and thus, we obtain
(3.1780) G μ ν ( 1 ) = R μ ν ( 1 ) 1 2 R ( 1 ) η μ ν = 1 2 ( h μ ν 1 2 h η μ ν ) = 1 2 h ¯ μ ν . (3.1780) G μ ν ( 1 ) = R μ ν ( 1 ) 1 2 R ( 1 ) η μ ν = 1 2 h μ ν 1 2 h η μ ν = 1 2 h ¯ μ ν . {:(3.1780)G_(mu nu)^((1))=R_(mu nu)^((1))-(1)/(2)R^((1))eta_(mu nu)=-(1)/(2)◻(h_(mu nu)-(1)/(2)heta_(mu nu))=-(1)/(2)◻ bar(h)_(mu nu).:}\begin{equation*} G_{\mu \nu}^{(1)}=R_{\mu \nu}^{(1)}-\frac{1}{2} R^{(1)} \eta_{\mu \nu}=-\frac{1}{2} \square\left(h_{\mu \nu}-\frac{1}{2} h \eta_{\mu \nu}\right)=-\frac{1}{2} \square \bar{h}_{\mu \nu} . \tag{3.1780} \end{equation*}(3.1780)Gμν(1)=Rμν(1)12R(1)ημν=12(hμν12hημν)=12h¯μν.
This means that the linearized Einstein equations turn into standard wave equations of the form
(3.1781) h ¯ μ ν = 16 π G c 4 T μ ν (3.1781) h ¯ μ ν = 16 π G c 4 T μ ν {:(3.1781)◻ bar(h)_(mu nu)=-(16 pi G)/(c^(4))T_(mu nu):}\begin{equation*} \square \bar{h}_{\mu \nu}=-\frac{16 \pi G}{c^{4}} T_{\mu \nu} \tag{3.1781} \end{equation*}(3.1781)h¯μν=16πGc4Tμν
2.113
a) Considering the metric corresponding to a weak gravitational potential Φ ( x ) Φ ( x ) Phi(x)\Phi(\mathbf{x})Φ(x) and setting c = 1 c = 1 c=1c=1c=1, we have the Lagrangian
(3.1782) L = [ 1 2 Φ ( x ) ] t ˙ 2 [ 1 + 2 Φ ( x ) ] ( x ˙ 2 + y ˙ 2 + z ˙ 2 ) (3.1782) L = [ 1 2 Φ ( x ) ] t ˙ 2 [ 1 + 2 Φ ( x ) ] x ˙ 2 + y ˙ 2 + z ˙ 2 {:(3.1782)L=[1-2Phi(x)]t^(˙)^(2)-[1+2Phi(x)](x^(˙)^(2)+y^(˙)^(2)+z^(˙)^(2)):}\begin{equation*} \mathcal{L}=[1-2 \Phi(\mathbf{x})] \dot{t}^{2}-[1+2 \Phi(\mathbf{x})]\left(\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2}\right) \tag{3.1782} \end{equation*}(3.1782)L=[12Φ(x)]t˙2[1+2Φ(x)](x˙2+y˙2+z˙2)
where Φ ( x ) Φ ( x ) Phi(x)\Phi(\mathbf{x})Φ(x) is independent of time t t ttt. Using the Euler-Lagrange equations, we find that
d d τ L t ˙ L t = 0 d d τ [ ( 1 2 Φ ( x ) ) t ˙ ] = 0 ( 1 2 Φ ( x ) ) t ˙ = d d τ L t ˙ L t = 0 d d τ [ ( 1 2 Φ ( x ) ) t ˙ ] = 0 ( 1 2 Φ ( x ) ) t ˙ = (d)/(d tau)(delL)/(del(t^(˙)))-(delL)/(del t)=0quad=>quad(d)/(d tau)[(1-2Phi(x))t^(˙)]=0quad=>quad(1-2Phi(x))t^(˙)=\frac{d}{d \tau} \frac{\partial \mathcal{L}}{\partial \dot{t}}-\frac{\partial \mathcal{L}}{\partial t}=0 \quad \Rightarrow \quad \frac{d}{d \tau}[(1-2 \Phi(\mathbf{x})) \dot{t}]=0 \quad \Rightarrow \quad(1-2 \Phi(\mathbf{x})) \dot{t}=ddτLt˙Lt=0ddτ[(12Φ(x))t˙]=0(12Φ(x))t˙= const.,
d d τ L x ˙ i L x i = 0 x ¨ i + 2 1 + 2 Φ ( x ) ( Φ x x ˙ + Φ y y ˙ + Φ z z ˙ ) x ˙ i (3.1784) Φ x i 1 + 2 Φ ( x ) ( z ˙ 2 + x ˙ 2 + y ˙ 2 + z ˙ 2 ) = 0 . d d τ L x ˙ i L x i = 0 x ¨ i + 2 1 + 2 Φ ( x ) Φ x x ˙ + Φ y y ˙ + Φ z z ˙ x ˙ i (3.1784) Φ x i 1 + 2 Φ ( x ) z ˙ 2 + x ˙ 2 + y ˙ 2 + z ˙ 2 = 0 . {:[(d)/(d tau)(delL)/(delx^(˙)^(i))-(delL)/(delx^(i))=0],[=>quadx^(¨)^(i)+(2)/(1+2Phi(x))(Phi_(x)^(')(x^(˙))+Phi_(y)^(')(y^(˙))+Phi_(z)^(')(z^(˙)))x^(˙)^(i)],[(3.1784)quad-(Phi_(x^(i))^('))/(1+2Phi(x))(z^(˙)^(2)+x^(˙)^(2)+y^(˙)^(2)+z^(˙)^(2))=0.]:}\begin{align*} & \frac{d}{d \tau} \frac{\partial \mathcal{L}}{\partial \dot{x}^{i}}-\frac{\partial \mathcal{L}}{\partial x^{i}}=0 \\ & \Rightarrow \quad \ddot{x}^{i}+\frac{2}{1+2 \Phi(\mathbf{x})}\left(\Phi_{x}^{\prime} \dot{x}+\Phi_{y}^{\prime} \dot{y}+\Phi_{z}^{\prime} \dot{z}\right) \dot{x}^{i} \\ & \quad-\frac{\Phi_{x^{i}}^{\prime}}{1+2 \Phi(\mathbf{x})}\left(\dot{z}^{2}+\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2}\right)=0 . \tag{3.1784} \end{align*}ddτLx˙iLxi=0x¨i+21+2Φ(x)(Φxx˙+Φyy˙+Φzz˙)x˙i(3.1784)Φxi1+2Φ(x)(z˙2+x˙2+y˙2+z˙2)=0.
However, since the metric only holds in the weak-field limit (where only lowestorder terms of Φ Φ Phi\PhiΦ are kept), we should write the Euler-Lagrange equations in this limit and have
(3.1785) ( 1 2 Φ ( x ) ) t ˙ = const., (3.1786) x ¨ i + 2 ( Φ x x ˙ + Φ y y ˙ + Φ z z ˙ ) x ˙ i Φ x i ( t ˙ 2 + x ˙ 2 + y ˙ 2 + z ˙ 2 ) = 0 . (3.1785) ( 1 2 Φ ( x ) ) t ˙ =  const.,  (3.1786) x ¨ i + 2 Φ x x ˙ + Φ y y ˙ + Φ z z ˙ x ˙ i Φ x i t ˙ 2 + x ˙ 2 + y ˙ 2 + z ˙ 2 = 0 . {:[(3.1785)(1-2Phi(x))t^(˙)=" const., "],[(3.1786)x^(¨)^(i)+2(Phi_(x)^(')(x^(˙))+Phi_(y)^(')(y^(˙))+Phi_(z)^(')(z^(˙)))x^(˙)^(i)-Phi_(x^(i))^(')(t^(˙)^(2)+x^(˙)^(2)+y^(˙)^(2)+z^(˙)^(2))=0.]:}\begin{gather*} (1-2 \Phi(\mathbf{x})) \dot{t}=\text { const., } \tag{3.1785}\\ \ddot{x}^{i}+2\left(\Phi_{x}^{\prime} \dot{x}+\Phi_{y}^{\prime} \dot{y}+\Phi_{z}^{\prime} \dot{z}\right) \dot{x}^{i}-\Phi_{x^{i}}^{\prime}\left(\dot{t}^{2}+\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2}\right)=0 . \tag{3.1786} \end{gather*}(3.1785)(12Φ(x))t˙= const., (3.1786)x¨i+2(Φxx˙+Φyy˙+Φzz˙)x˙iΦxi(t˙2+x˙2+y˙2+z˙2)=0.
Furthermore, we apply the nonrelativistic limit, which says that x ˙ i t ˙ x ˙ i t ˙ x^(˙)^(i)≪t^(˙)\dot{x}^{i} \ll \dot{t}x˙it˙, since it holds for a slowly moving massive particle that τ t τ t tau~~t\tau \approx tτt and d x i / d t 1 d x i / d t 1 dx^(i)//dt≪1d x^{i} / d t \ll 1dxi/dt1, which implies that x ˙ i t ˙ 1 x ˙ i t ˙ 1 x^(˙)^(i)≪t^(˙)~~1\dot{x}^{i} \ll \dot{t} \approx 1x˙it˙1. Thus, we obtain the simple equations
(3.1787) t ¨ 0 , x ¨ i Φ x i 0 , (3.1787) t ¨ 0 , x ¨ i Φ x i 0 , {:(3.1787)t^(¨)≃0","quadx^(¨)^(i)-Phi_(x^(i))^(')≃0",":}\begin{equation*} \ddot{t} \simeq 0, \quad \ddot{x}^{i}-\Phi_{x^{i}}^{\prime} \simeq 0, \tag{3.1787} \end{equation*}(3.1787)t¨0,x¨iΦxi0,
which are the geodesic equations in the nonrelativistic and weak-field limits. The first equation means that energy (or t ˙ t ˙ t^(˙)\dot{t}t˙ ) is conserved (the integration constant is chosen such that t ˙ = 1 t ˙ = 1 t^(˙)=1\dot{t}=1t˙=1, which then coincides with L = 1 L = 1 L=1\mathcal{L}=1L=1 in the nonrelativistic limit), whereas multiplying the second equation with the mass m m mmm of the massive test particle leads to
(3.1788) m x ¨ i = m Φ x i = m Φ x i = m i Φ = m i Φ (3.1788) m x ¨ i = m Φ x i = m Φ x i = m i Φ = m i Φ {:(3.1788)mx^(¨)^(i)=mPhi_(x^(i))^(')=m(del Phi)/(delx^(i))=mdel_(i)Phi=-mdel^(i)Phi:}\begin{equation*} m \ddot{x}^{i}=m \Phi_{x^{i}}^{\prime}=m \frac{\partial \Phi}{\partial x^{i}}=m \partial_{i} \Phi=-m \partial^{i} \Phi \tag{3.1788} \end{equation*}(3.1788)mx¨i=mΦxi=mΦxi=miΦ=miΦ
which is just Newton's second law, i.e., m a = F = m Φ ( x ) m a = F = m Φ ( x ) ma=F=-m grad Phi(x)m \mathbf{a}=\mathbf{F}=-m \nabla \Phi(\mathbf{x})ma=F=mΦ(x), where the force F = m Φ ( x ) F = m Φ ( x ) F=-m grad Phi(x)\mathbf{F}=-m \nabla \Phi(\mathbf{x})F=mΦ(x) is that of a weak gravitational potential Φ ( x ) Φ ( x ) Phi(x)\Phi(\mathbf{x})Φ(x). Compare the solution to Problem 2.111 a).
b) Using the metric for the weak gravitational potential Φ ( x ) Φ ( x ) Phi(x)\Phi(\mathbf{x})Φ(x), we have
(3.1789) g 00 ( x ) = [ c 2 2 Φ ( x ) ] 1 c 2 = 1 2 c 2 Φ ( x ) (3.1789) g 00 ( x ) = c 2 2 Φ ( x ) 1 c 2 = 1 2 c 2 Φ ( x ) {:(3.1789)g_(00)(x)=[c^(2)-2Phi(x)](1)/(c^(2))=1-(2)/(c^(2))Phi(x):}\begin{equation*} g_{00}(\mathbf{x})=\left[c^{2}-2 \Phi(\mathbf{x})\right] \frac{1}{c^{2}}=1-\frac{2}{c^{2}} \Phi(\mathbf{x}) \tag{3.1789} \end{equation*}(3.1789)g00(x)=[c22Φ(x)]1c2=12c2Φ(x)
Now, assuming Φ ( x ) = g z Φ ( x ) = g z Phi(x)=-gz\Phi(\mathbf{x})=-g zΦ(x)=gz, we find that
(3.1790) g 00 ( z ) = 1 + 2 c 2 g z (3.1790) g 00 ( z ) = 1 + 2 c 2 g z {:(3.1790)g_(00)(z)=1+(2)/(c^(2))gz:}\begin{equation*} g_{00}(z)=1+\frac{2}{c^{2}} g z \tag{3.1790} \end{equation*}(3.1790)g00(z)=1+2c2gz
Then, using the formula for the gravitational redshift to compute the photon's gravitational redshift, we obtain
ω ω = 1 + 2 c 2 g 0 1 + 2 c 2 g h 1 g h c 2 (3.1791) z = λ λ 1 = ω ω 1 = 1 + 2 g h c 2 1 g h c 2 ω ω = 1 + 2 c 2 g 0 1 + 2 c 2 g h 1 g h c 2 (3.1791) z = λ λ 1 = ω ω 1 = 1 + 2 g h c 2 1 g h c 2 {:[(omega^('))/(omega)=sqrt((1+(2)/(c^(2))g*0)/(1+(2)/(c^(2))g*h))≃1-(gh)/(c^(2))],[(3.1791)quad=>quad z=(lambda^('))/(lambda)-1=(omega)/(omega^('))-1=sqrt(1+(2gh)/(c^(2)))-1≃(gh)/(c^(2))]:}\begin{align*} & \frac{\omega^{\prime}}{\omega}=\sqrt{\frac{1+\frac{2}{c^{2}} g \cdot 0}{1+\frac{2}{c^{2}} g \cdot h}} \simeq 1-\frac{g h}{c^{2}} \\ & \quad \Rightarrow \quad z=\frac{\lambda^{\prime}}{\lambda}-1=\frac{\omega}{\omega^{\prime}}-1=\sqrt{1+\frac{2 g h}{c^{2}}}-1 \simeq \frac{g h}{c^{2}} \tag{3.1791} \end{align*}ωω=1+2c2g01+2c2gh1ghc2(3.1791)z=λλ1=ωω1=1+2ghc21ghc2
where h > 0 h > 0 h > 0h>0h>0. Thus, a photon emitted at a lower gravitational potential with an angular frequency ω ω omega\omegaω is received at a higher gravitational potential with a smaller angular frequency ω ω omega^(')\omega^{\prime}ω, i.e., it is redshifted, although the emitter and the receiver are not in relative motion.
Finally, consider a particular frame to be at rest when the photon starts its upward climb in the gravitational potential and falls freely after that. Since the photon climbs from z = 0 z = 0 z=0z=0z=0 to z = h > 0 z = h > 0 z=h > 0z=h>0z=h>0, i.e., a distance h h hhh, it takes the time Δ t = h / c Δ t = h / c Delta t=h//c\Delta t=h / cΔt=h/c to arrive at the higher gravitational potential. During this time, the frame has obtained the velocity v = g Δ t = g h / c v = g Δ t = g h / c v=g Delta t=gh//cv=g \Delta t=g h / cv=gΔt=gh/c downward relative to the higher gravitational potential. Therefore, using formula for the Doppler shift, the photon's angular frequency ω ff ω ff omega_(ff)\omega_{\mathrm{ff}}ωff relative to the freely falling frame is
(3.1792) ω ff ω = 1 ( g h / c 2 ) 1 + ( g h / c 2 ) 1 + g h c 2 (3.1792) ω ff ω = 1 g h / c 2 1 + g h / c 2 1 + g h c 2 {:(3.1792)(omega_(ff))/(omega^('))=sqrt((1-(-gh//c^(2)))/(1+(-gh//c^(2))))≃1+(gh)/(c^(2)):}\begin{equation*} \frac{\omega_{\mathrm{ff}}}{\omega^{\prime}}=\sqrt{\frac{1-\left(-g h / c^{2}\right)}{1+\left(-g h / c^{2}\right)}} \simeq 1+\frac{g h}{c^{2}} \tag{3.1792} \end{equation*}(3.1792)ωffω=1(gh/c2)1+(gh/c2)1+ghc2
which means that
(3.1793) ω ff ω ( 1 + g h c 2 ) ω ( 1 g h c 2 ) ( 1 + g h c 2 ) ω . (3.1793) ω ff ω 1 + g h c 2 ω 1 g h c 2 1 + g h c 2 ω . {:(3.1793)omega_(ff)≃omega^(')(1+(gh)/(c^(2)))≃omega(1-(gh)/(c^(2)))(1+(gh)/(c^(2)))≃omega.:}\begin{equation*} \omega_{\mathrm{ff}} \simeq \omega^{\prime}\left(1+\frac{g h}{c^{2}}\right) \simeq \omega\left(1-\frac{g h}{c^{2}}\right)\left(1+\frac{g h}{c^{2}}\right) \simeq \omega . \tag{3.1793} \end{equation*}(3.1793)ωffω(1+ghc2)ω(1ghc2)(1+ghc2)ω.
Thus, there is no redshift in a freely falling frame, confirming that it is a local inertial frame. This is the essence of Einstein's equivalence principle, which means that the gravitational field can be transformed away by transforming to an appropriate accelerating ("freely falling") frame of reference.

2.114

a) Consider two massive particles moving freely on two close paths, where one particle has the spacetime path x μ ( τ ) x μ ( τ ) x^(mu)(tau)x^{\mu}(\tau)xμ(τ) and the other one has the spacetime path x μ ( τ ) + s μ ( τ ) x μ ( τ ) + s μ ( τ ) x^(mu)(tau)+s^(mu)(tau)x^{\mu}(\tau)+s^{\mu}(\tau)xμ(τ)+sμ(τ). Therefore, the two particles are separated by the displacement vector s μ ( τ ) s μ ( τ ) s^(mu)(tau)s^{\mu}(\tau)sμ(τ), which is small. The geodesic equations (or the equations of motion) for the two particles are given by
(3.1794) x ¨ μ + Γ α β μ ( x ) x ˙ α x ˙ β = 0 , ( x ¨ μ + s ¨ μ ) + Γ α β μ ( x + s ) ( x ˙ α + s ˙ α ) ( x ˙ β + s ˙ β ) = 0 . (3.1794) x ¨ μ + Γ α β μ ( x ) x ˙ α x ˙ β = 0 , x ¨ μ + s ¨ μ + Γ α β μ ( x + s ) x ˙ α + s ˙ α x ˙ β + s ˙ β = 0 . {:(3.1794)x^(¨)^(mu)+Gamma_(alpha beta)^(mu)(x)x^(˙)^(alpha)x^(˙)^(beta)=0","quad(x^(¨)^(mu)+s^(¨)^(mu))+Gamma_(alpha beta)^(mu)(x+s)(x^(˙)^(alpha)+s^(˙)^(alpha))(x^(˙)^(beta)+s^(˙)^(beta))=0.:}\begin{equation*} \ddot{x}^{\mu}+\Gamma_{\alpha \beta}^{\mu}(x) \dot{x}^{\alpha} \dot{x}^{\beta}=0, \quad\left(\ddot{x}^{\mu}+\ddot{s}^{\mu}\right)+\Gamma_{\alpha \beta}^{\mu}(x+s)\left(\dot{x}^{\alpha}+\dot{s}^{\alpha}\right)\left(\dot{x}^{\beta}+\dot{s}^{\beta}\right)=0 . \tag{3.1794} \end{equation*}(3.1794)x¨μ+Γαβμ(x)x˙αx˙β=0,(x¨μ+s¨μ)+Γαβμ(x+s)(x˙α+s˙α)(x˙β+s˙β)=0.
Since s μ s μ s^(mu)s^{\mu}sμ is small, we can series expand the Christoffel symbols Γ α β μ ( x + s ) Γ α β μ ( x + s ) Gamma_(alpha beta)^(mu)(x+s)\Gamma_{\alpha \beta}^{\mu}(x+s)Γαβμ(x+s) in the second equation as
(3.1795) Γ α β μ ( x + s ) Γ α β μ ( x ) + λ Γ α β μ ( x ) s λ (3.1795) Γ α β μ ( x + s ) Γ α β μ ( x ) + λ Γ α β μ ( x ) s λ {:(3.1795)Gamma_(alpha beta)^(mu)(x+s)≃Gamma_(alpha beta)^(mu)(x)+del_(lambda)Gamma_(alpha beta)^(mu)(x)s^(lambda):}\begin{equation*} \Gamma_{\alpha \beta}^{\mu}(x+s) \simeq \Gamma_{\alpha \beta}^{\mu}(x)+\partial_{\lambda} \Gamma_{\alpha \beta}^{\mu}(x) s^{\lambda} \tag{3.1795} \end{equation*}(3.1795)Γαβμ(x+s)Γαβμ(x)+λΓαβμ(x)sλ
Thus, from the difference between the two geodesics equations, we have to first order in s s sss
(3.1796) s ¨ μ 2 Γ α β μ ( x ) s ˙ α x ˙ β λ Γ α β μ ( x ) s λ x ˙ α x ˙ β , (3.1796) s ¨ μ 2 Γ α β μ ( x ) s ˙ α x ˙ β λ Γ α β μ ( x ) s λ x ˙ α x ˙ β , {:(3.1796)s^(¨)^(mu)≃-2Gamma_(alpha beta)^(mu)(x)s^(˙)^(alpha)x^(˙)^(beta)-del_(lambda)Gamma_(alpha beta)^(mu)(x)s^(lambda)x^(˙)^(alpha)x^(˙)^(beta)",":}\begin{equation*} \ddot{s}^{\mu} \simeq-2 \Gamma_{\alpha \beta}^{\mu}(x) \dot{s}^{\alpha} \dot{x}^{\beta}-\partial_{\lambda} \Gamma_{\alpha \beta}^{\mu}(x) s^{\lambda} \dot{x}^{\alpha} \dot{x}^{\beta}, \tag{3.1796} \end{equation*}(3.1796)s¨μ2Γαβμ(x)s˙αx˙βλΓαβμ(x)sλx˙αx˙β,
where we used Γ α β μ ( x ) = Γ β α μ ( x ) Γ α β μ ( x ) = Γ β α μ ( x ) Gamma_(alpha beta)^(mu)(x)=Gamma_(beta alpha)^(mu)(x)\Gamma_{\alpha \beta}^{\mu}(x)=\Gamma_{\beta \alpha}^{\mu}(x)Γαβμ(x)=Γβαμ(x). Using the definition of the parallel transport of a vector V μ V μ V^(mu)V^{\mu}Vμ along a path x μ ( τ ) x μ ( τ ) x^(mu)(tau)x^{\mu}(\tau)xμ(τ), i.e.,
(3.1797) D V μ D τ = V ˙ μ + Γ α β μ V α x ˙ β = 0 (3.1797) D V μ D τ = V ˙ μ + Γ α β μ V α x ˙ β = 0 {:(3.1797)(DV^(mu))/(D tau)=V^(˙)^(mu)+Gamma_(alpha beta)^(mu)V^(alpha)x^(˙)^(beta)=0:}\begin{equation*} \frac{D V^{\mu}}{D \tau}=\dot{V}^{\mu}+\Gamma_{\alpha \beta}^{\mu} V^{\alpha} \dot{x}^{\beta}=0 \tag{3.1797} \end{equation*}(3.1797)DVμDτ=V˙μ+ΓαβμVαx˙β=0
we have directly the first derivative of s μ s μ s^(mu)s^{\mu}sμ
(3.1798) D s μ D τ = V ˙ μ + Γ α β μ V α x ˙ β (3.1798) D s μ D τ = V ˙ μ + Γ α β μ V α x ˙ β {:(3.1798)(Ds^(mu))/(D tau)=V^(˙)^(mu)+Gamma_(alpha beta)^(mu)V^(alpha)x^(˙)^(beta):}\begin{equation*} \frac{D s^{\mu}}{D \tau}=\dot{V}^{\mu}+\Gamma_{\alpha \beta}^{\mu} V^{\alpha} \dot{x}^{\beta} \tag{3.1798} \end{equation*}(3.1798)DsμDτ=V˙μ+ΓαβμVαx˙β
and we can also find the second derivative of s μ s μ s^(mu)s^{\mu}sμ as
D 2 s μ D τ 2 = D D τ D s μ D τ = d d τ D s μ D τ + Γ α β μ D s α D τ x ˙ β = d d τ ( s ˙ μ + Γ α β μ s α x ˙ β ) + Γ α β μ ( s ˙ α + Γ λ ρ α s λ x ˙ ρ ) x ˙ β (3.1799) = s ¨ μ + λ Γ α β μ x ˙ λ s α x ˙ β + Γ α β μ s ˙ α x ˙ β + Γ α β μ s α x ¨ β + Γ α β μ s ˙ α x ˙ β + Γ α β μ Γ λ ρ α s λ x ˙ ρ x ˙ β D 2 s μ D τ 2 = D D τ D s μ D τ = d d τ D s μ D τ + Γ α β μ D s α D τ x ˙ β = d d τ s ˙ μ + Γ α β μ s α x ˙ β + Γ α β μ s ˙ α + Γ λ ρ α s λ x ˙ ρ x ˙ β (3.1799) = s ¨ μ + λ Γ α β μ x ˙ λ s α x ˙ β + Γ α β μ s ˙ α x ˙ β + Γ α β μ s α x ¨ β + Γ α β μ s ˙ α x ˙ β + Γ α β μ Γ λ ρ α s λ x ˙ ρ x ˙ β {:[(D^(2)s^(mu))/(Dtau^(2))=(D)/(D tau)(Ds^(mu))/(D tau)=(d)/(d tau)(Ds^(mu))/(D tau)+Gamma_(alpha beta)^(mu)(Ds^(alpha))/(D tau)x^(˙)^(beta)],[=(d)/(d tau)(s^(˙)^(mu)+Gamma_(alpha beta)^(mu)s^(alpha)x^(˙)^(beta))+Gamma_(alpha beta)^(mu)(s^(˙)^(alpha)+Gamma_(lambda rho)^(alpha)s^(lambda)x^(˙)^(rho))x^(˙)^(beta)],[(3.1799)=s^(¨)^(mu)+del_(lambda)Gamma_(alpha beta)^(mu)x^(˙)^(lambda)s^(alpha)x^(˙)^(beta)+Gamma_(alpha beta)^(mu)s^(˙)^(alpha)x^(˙)^(beta)+Gamma_(alpha beta)^(mu)s^(alpha)x^(¨)^(beta)+Gamma_(alpha beta)^(mu)s^(˙)^(alpha)x^(˙)^(beta)+Gamma_(alpha beta)^(mu)Gamma_(lambda rho)^(alpha)s^(lambda)x^(˙)^(rho)x^(˙)^(beta)]:}\begin{align*} \frac{D^{2} s^{\mu}}{D \tau^{2}} & =\frac{D}{D \tau} \frac{D s^{\mu}}{D \tau}=\frac{d}{d \tau} \frac{D s^{\mu}}{D \tau}+\Gamma_{\alpha \beta}^{\mu} \frac{D s^{\alpha}}{D \tau} \dot{x}^{\beta} \\ & =\frac{d}{d \tau}\left(\dot{s}^{\mu}+\Gamma_{\alpha \beta}^{\mu} s^{\alpha} \dot{x}^{\beta}\right)+\Gamma_{\alpha \beta}^{\mu}\left(\dot{s}^{\alpha}+\Gamma_{\lambda \rho}^{\alpha} s^{\lambda} \dot{x}^{\rho}\right) \dot{x}^{\beta} \\ & =\ddot{s}^{\mu}+\partial_{\lambda} \Gamma_{\alpha \beta}^{\mu} \dot{x}^{\lambda} s^{\alpha} \dot{x}^{\beta}+\Gamma_{\alpha \beta}^{\mu} \dot{s}^{\alpha} \dot{x}^{\beta}+\Gamma_{\alpha \beta}^{\mu} s^{\alpha} \ddot{x}^{\beta}+\Gamma_{\alpha \beta}^{\mu} \dot{s}^{\alpha} \dot{x}^{\beta}+\Gamma_{\alpha \beta}^{\mu} \Gamma_{\lambda \rho}^{\alpha} s^{\lambda} \dot{x}^{\rho} \dot{x}^{\beta} \tag{3.1799} \end{align*}D2sμDτ2=DDτDsμDτ=ddτDsμDτ+ΓαβμDsαDτx˙β=ddτ(s˙μ+Γαβμsαx˙β)+Γαβμ(s˙α+Γλραsλx˙ρ)x˙β(3.1799)=s¨μ+λΓαβμx˙λsαx˙β+Γαβμs˙αx˙β+Γαβμsαx¨β+Γαβμs˙αx˙β+ΓαβμΓλραsλx˙ρx˙β
In order to simplify the second derivative, we use the above-derived expressions for s ¨ μ s ¨ μ s^(¨)^(mu)\ddot{s}^{\mu}s¨μ and x ¨ β x ¨ β x^(¨)^(beta)\ddot{x}^{\beta}x¨β and obtain
D 2 s μ D τ 2 = 2 Γ α β μ s ˙ α x ˙ β λ Γ α β μ s λ x ˙ α x ˙ β + λ Γ α β μ x ˙ λ s α x ˙ β + Γ α β μ s ˙ α x ˙ β Γ α β μ Γ λ ρ β s α x ˙ λ x ˙ ρ + Γ α β μ s ˙ α x ˙ β + Γ α β μ Γ λ ρ α s λ x ˙ ρ x ˙ β = λ Γ α β μ s λ x ˙ α x ˙ β + λ Γ α β μ x ˙ λ s α x ˙ β Γ α β μ Γ λ ρ β s α x ˙ λ x ˙ ρ + Γ α β μ Γ λ ρ α s λ x ˙ ρ x ˙ β (3.1800) = ( λ Γ α β μ α Γ λ β μ + Γ λ ρ μ Γ α β ρ Γ ρ β μ Γ λ α ρ ) s λ x ˙ α x ˙ β D 2 s μ D τ 2 = 2 Γ α β μ s ˙ α x ˙ β λ Γ α β μ s λ x ˙ α x ˙ β + λ Γ α β μ x ˙ λ s α x ˙ β + Γ α β μ s ˙ α x ˙ β Γ α β μ Γ λ ρ β s α x ˙ λ x ˙ ρ + Γ α β μ s ˙ α x ˙ β + Γ α β μ Γ λ ρ α s λ x ˙ ρ x ˙ β = λ Γ α β μ s λ x ˙ α x ˙ β + λ Γ α β μ x ˙ λ s α x ˙ β Γ α β μ Γ λ ρ β s α x ˙ λ x ˙ ρ + Γ α β μ Γ λ ρ α s λ x ˙ ρ x ˙ β (3.1800) = λ Γ α β μ α Γ λ β μ + Γ λ ρ μ Γ α β ρ Γ ρ β μ Γ λ α ρ s λ x ˙ α x ˙ β {:[(D^(2)s^(mu))/(Dtau^(2))=-2Gamma_(alpha beta)^(mu)s^(˙)^(alpha)x^(˙)^(beta)-del_(lambda)Gamma_(alpha beta)^(mu)s^(lambda)x^(˙)^(alpha)x^(˙)^(beta)+del_(lambda)Gamma_(alpha beta)^(mu)x^(˙)^(lambda)s^(alpha)x^(˙)^(beta)+Gamma_(alpha beta)^(mu)s^(˙)^(alpha)x^(˙)^(beta)],[-Gamma_(alpha beta)^(mu)Gamma_(lambda rho)^(beta)s^(alpha)x^(˙)^(lambda)x^(˙)^(rho)+Gamma_(alpha beta)^(mu)s^(˙)^(alpha)x^(˙)^(beta)+Gamma_(alpha beta)^(mu)Gamma_(lambda rho)^(alpha)s^(lambda)x^(˙)^(rho)x^(˙)^(beta)],[=-del_(lambda)Gamma_(alpha beta)^(mu)s^(lambda)x^(˙)^(alpha)x^(˙)^(beta)+del_(lambda)Gamma_(alpha beta)^(mu)x^(˙)^(lambda)s^(alpha)x^(˙)^(beta)-Gamma_(alpha beta)^(mu)Gamma_(lambda rho)^(beta)s^(alpha)x^(˙)^(lambda)x^(˙)^(rho)+Gamma_(alpha beta)^(mu)Gamma_(lambda rho)^(alpha)s^(lambda)x^(˙)^(rho)x^(˙)^(beta)],[(3.1800)=-(del_(lambda)Gamma_(alpha beta)^(mu)-del_(alpha)Gamma_(lambda beta)^(mu)+Gamma_(lambda rho)^(mu)Gamma_(alpha beta)^(rho)-Gamma_(rho beta)^(mu)Gamma_(lambda alpha)^(rho))s^(lambda)x^(˙)^(alpha)x^(˙)^(beta)]:}\begin{align*} \frac{D^{2} s^{\mu}}{D \tau^{2}}= & -2 \Gamma_{\alpha \beta}^{\mu} \dot{s}^{\alpha} \dot{x}^{\beta}-\partial_{\lambda} \Gamma_{\alpha \beta}^{\mu} s^{\lambda} \dot{x}^{\alpha} \dot{x}^{\beta}+\partial_{\lambda} \Gamma_{\alpha \beta}^{\mu} \dot{x}^{\lambda} s^{\alpha} \dot{x}^{\beta}+\Gamma_{\alpha \beta}^{\mu} \dot{s}^{\alpha} \dot{x}^{\beta} \\ & -\Gamma_{\alpha \beta}^{\mu} \Gamma_{\lambda \rho}^{\beta} s^{\alpha} \dot{x}^{\lambda} \dot{x}^{\rho}+\Gamma_{\alpha \beta}^{\mu} \dot{s}^{\alpha} \dot{x}^{\beta}+\Gamma_{\alpha \beta}^{\mu} \Gamma_{\lambda \rho}^{\alpha} s^{\lambda} \dot{x}^{\rho} \dot{x}^{\beta} \\ = & -\partial_{\lambda} \Gamma_{\alpha \beta}^{\mu} s^{\lambda} \dot{x}^{\alpha} \dot{x}^{\beta}+\partial_{\lambda} \Gamma_{\alpha \beta}^{\mu} \dot{x}^{\lambda} s^{\alpha} \dot{x}^{\beta}-\Gamma_{\alpha \beta}^{\mu} \Gamma_{\lambda \rho}^{\beta} s^{\alpha} \dot{x}^{\lambda} \dot{x}^{\rho}+\Gamma_{\alpha \beta}^{\mu} \Gamma_{\lambda \rho}^{\alpha} s^{\lambda} \dot{x}^{\rho} \dot{x}^{\beta} \\ = & -\left(\partial_{\lambda} \Gamma_{\alpha \beta}^{\mu}-\partial_{\alpha} \Gamma_{\lambda \beta}^{\mu}+\Gamma_{\lambda \rho}^{\mu} \Gamma_{\alpha \beta}^{\rho}-\Gamma_{\rho \beta}^{\mu} \Gamma_{\lambda \alpha}^{\rho}\right) s^{\lambda} \dot{x}^{\alpha} \dot{x}^{\beta} \tag{3.1800} \end{align*}D2sμDτ2=2Γαβμs˙αx˙βλΓαβμsλx˙αx˙β+λΓαβμx˙λsαx˙β+Γαβμs˙αx˙βΓαβμΓλρβsαx˙λx˙ρ+Γαβμs˙αx˙β+ΓαβμΓλραsλx˙ρx˙β=λΓαβμsλx˙αx˙β+λΓαβμx˙λsαx˙βΓαβμΓλρβsαx˙λx˙ρ+ΓαβμΓλραsλx˙ρx˙β(3.1800)=(λΓαβμαΓλβμ+ΓλρμΓαβρΓρβμΓλαρ)sλx˙αx˙β
where in the last step we relabeled some of the summation indices. It holds that α Γ λ β μ s λ x ˙ α x ˙ β = β Γ λ α μ s λ x ˙ α x ˙ β α Γ λ β μ s λ x ˙ α x ˙ β = β Γ λ α μ s λ x ˙ α x ˙ β del_(alpha)Gamma_(lambda beta)^(mu)s^(lambda)x^(˙)^(alpha)x^(˙)^(beta)=del_(beta)Gamma_(lambda alpha)^(mu)s^(lambda)x^(˙)^(alpha)x^(˙)^(beta)\partial_{\alpha} \Gamma_{\lambda \beta}^{\mu} s^{\lambda} \dot{x}^{\alpha} \dot{x}^{\beta}=\partial_{\beta} \Gamma_{\lambda \alpha}^{\mu} s^{\lambda} \dot{x}^{\alpha} \dot{x}^{\beta}αΓλβμsλx˙αx˙β=βΓλαμsλx˙αx˙β and that Γ α β μ = Γ β α μ Γ α β μ = Γ β α μ Gamma_(alpha beta)^(mu)=Gamma_(beta alpha)^(mu)\Gamma_{\alpha \beta}^{\mu}=\Gamma_{\beta \alpha}^{\mu}Γαβμ=Γβαμ. Thus, we can identify the Riemann curvature tensor as
(3.1801) R α λ β μ = λ Γ β α μ β Γ λ α μ + Γ λ ρ μ Γ β α ρ Γ β ρ μ Γ λ α ρ (3.1801) R α λ β μ = λ Γ β α μ β Γ λ α μ + Γ λ ρ μ Γ β α ρ Γ β ρ μ Γ λ α ρ {:(3.1801)R_(alpha lambda beta)^(mu)=del_(lambda)Gamma_(beta alpha)^(mu)-del_(beta)Gamma_(lambda alpha)^(mu)+Gamma_(lambda rho)^(mu)Gamma_(beta alpha)^(rho)-Gamma_(beta rho)^(mu)Gamma_(lambda alpha)^(rho):}\begin{equation*} R_{\alpha \lambda \beta}^{\mu}=\partial_{\lambda} \Gamma_{\beta \alpha}^{\mu}-\partial_{\beta} \Gamma_{\lambda \alpha}^{\mu}+\Gamma_{\lambda \rho}^{\mu} \Gamma_{\beta \alpha}^{\rho}-\Gamma_{\beta \rho}^{\mu} \Gamma_{\lambda \alpha}^{\rho} \tag{3.1801} \end{equation*}(3.1801)Rαλβμ=λΓβαμβΓλαμ+ΓλρμΓβαρΓβρμΓλαρ
and therefore, we have derived the equation of geodesic deviation, namely
(3.1802) D 2 s μ D τ 2 = R α λ β μ s λ x ˙ α x ˙ β = R α ν β μ s ν d x α d τ d x β d τ (3.1802) D 2 s μ D τ 2 = R α λ β μ s λ x ˙ α x ˙ β = R α ν β μ s ν d x α d τ d x β d τ {:(3.1802)(D^(2)s^(mu))/(Dtau^(2))=-R_(alpha lambda beta)^(mu)s^(lambda)x^(˙)^(alpha)x^(˙)^(beta)=-R_(alpha nu beta)^(mu)s^(nu)(dx^(alpha))/(d tau)(dx^(beta))/(d tau):}\begin{equation*} \frac{D^{2} s^{\mu}}{D \tau^{2}}=-R_{\alpha \lambda \beta}^{\mu} s^{\lambda} \dot{x}^{\alpha} \dot{x}^{\beta}=-R_{\alpha \nu \beta}^{\mu} s^{\nu} \frac{d x^{\alpha}}{d \tau} \frac{d x^{\beta}}{d \tau} \tag{3.1802} \end{equation*}(3.1802)D2sμDτ2=Rαλβμsλx˙αx˙β=Rανβμsνdxαdτdxβdτ
b) In the Newtonian limit, we can approximate the metric by g μ ν = η μ ν + g μ ν = η μ ν + g_(mu nu)=eta_(mu nu)+g_{\mu \nu}=\eta_{\mu \nu}+gμν=ημν+ h μ ν h μ ν h_(mu nu)h_{\mu \nu}hμν, where | h μ ν | 1 h μ ν 1 |h_(mu nu)|≪1\left|h_{\mu \nu}\right| \ll 1|hμν|1 is small, and therefore, we have the metric tensor g μ ν g μ ν g_(mu nu)g_{\mu \nu}gμν, the Christoffel symbols Γ μ ν λ Γ μ ν λ Gamma_(mu nu)^(lambda)\Gamma_{\mu \nu}^{\lambda}Γμνλ, and the Riemann curvature tensor R μ ν λ ρ R μ ν λ ρ R^(mu)_(nu lambda rho)R^{\mu}{ }_{\nu \lambda \rho}Rμνλρ as
(3.1803) g μ ν η μ ν + h μ v , Γ μ ν λ = 1 2 g λ ρ ( μ g ν ρ + ν g μ ρ ρ g μ ν ) (3.1804) 1 2 η λ ρ ( μ h v ρ + v h μ ρ ρ h μ ν ) Γ μ ν ( 1 ) λ , R ν λ ρ μ = λ Γ ρ v μ ρ Γ λ ν μ + Γ λ σ μ Γ ρ v σ Γ ρ σ μ Γ λ ν σ λ Γ ρ ν ( 1 ) μ ρ Γ λ ν ( 1 ) μ = 1 2 η μ σ ( λ ρ h v σ + v λ h ρ σ λ σ h ρ v λ ρ h v σ ν ρ h λ σ + ρ σ h λ ν ) = 1 2 ( v λ h ρ μ + μ ρ h λ ν v ρ h λ μ μ λ h ρ v ) (3.1805) = 1 2 η μ σ ( ν λ h ρ σ + σ ρ h v λ ν ρ h λ σ σ λ h v ρ ) R ν λ ρ ( 1 ) μ (3.1803) g μ ν η μ ν + h μ v , Γ μ ν λ = 1 2 g λ ρ μ g ν ρ + ν g μ ρ ρ g μ ν (3.1804) 1 2 η λ ρ μ h v ρ + v h μ ρ ρ h μ ν Γ μ ν ( 1 ) λ , R ν λ ρ μ = λ Γ ρ v μ ρ Γ λ ν μ + Γ λ σ μ Γ ρ v σ Γ ρ σ μ Γ λ ν σ λ Γ ρ ν ( 1 ) μ ρ Γ λ ν ( 1 ) μ = 1 2 η μ σ λ ρ h v σ + v λ h ρ σ λ σ h ρ v λ ρ h v σ ν ρ h λ σ + ρ σ h λ ν = 1 2 v λ h ρ μ + μ ρ h λ ν v ρ h λ μ μ λ h ρ v (3.1805) = 1 2 η μ σ ν λ h ρ σ + σ ρ h v λ ν ρ h λ σ σ λ h v ρ R ν λ ρ ( 1 ) μ {:[(3.1803)g_(mu nu)≃eta_(mu nu)+h_(mu v)","],[Gamma_(mu nu)^(lambda)=(1)/(2)g^(lambda rho)(del_(mu)g_(nu rho)+del_(nu)g_(mu rho)-del_(rho)g_(mu nu))],[(3.1804)≃(1)/(2)eta^(lambda rho)(del_(mu)h_(v rho)+del_(v)h_(mu rho)-del_(rho)h_(mu nu))-=Gamma_(mu nu)^((1)^(lambda))","],[R_(nu lambda rho)^(mu)=del_(lambda)Gamma_(rho v)^(mu)-del_(rho)Gamma_(lambda nu)^(mu)+Gamma_(lambda sigma)^(mu)Gamma_(rho v)^(sigma)-Gamma_(rho sigma)^(mu)Gamma_(lambda nu)^(sigma)≃del_(lambda)Gamma_(rho nu)^((1)mu)-del_(rho)Gamma_(lambda nu)^((1)mu)],[=(1)/(2)eta^(mu sigma)(del_(lambda)del_(rho)h_(v sigma)+del_(v)del_(lambda)h_(rho sigma)-del_(lambda)del_(sigma)h_(rho v)-del_(lambda)del_(rho)h_(v sigma)-del_(nu)del_(rho)h_(lambda sigma)+del_(rho)del_(sigma)h_(lambda nu))],[=(1)/(2)(del_(v)del_(lambda)h_(rho)^(mu)+del^(mu)del_(rho)h_(lambda nu)-del_(v)del_(rho)h_(lambda)^(mu)-del^(mu)del_(lambda)h_(rho v))],[(3.1805)=(1)/(2)eta^(mu sigma)(del_(nu)del_(lambda)h_(rho sigma)+del_(sigma)del_(rho)h_(v lambda)-del_(nu)del_(rho)h_(lambda sigma)-del_(sigma)del_(lambda)h_(v rho))-=R_(nu lambda rho)^((1)^(mu))]:}\begin{align*} g_{\mu \nu} & \simeq \eta_{\mu \nu}+h_{\mu v}, \tag{3.1803}\\ \Gamma_{\mu \nu}^{\lambda} & =\frac{1}{2} g^{\lambda \rho}\left(\partial_{\mu} g_{\nu \rho}+\partial_{\nu} g_{\mu \rho}-\partial_{\rho} g_{\mu \nu}\right) \\ & \simeq \frac{1}{2} \eta^{\lambda \rho}\left(\partial_{\mu} h_{v \rho}+\partial_{v} h_{\mu \rho}-\partial_{\rho} h_{\mu \nu}\right) \equiv \Gamma_{\mu \nu}^{(1)^{\lambda}}, \tag{3.1804}\\ R_{\nu \lambda \rho}^{\mu} & =\partial_{\lambda} \Gamma_{\rho v}^{\mu}-\partial_{\rho} \Gamma_{\lambda \nu}^{\mu}+\Gamma_{\lambda \sigma}^{\mu} \Gamma_{\rho v}^{\sigma}-\Gamma_{\rho \sigma}^{\mu} \Gamma_{\lambda \nu}^{\sigma} \simeq \partial_{\lambda} \Gamma_{\rho \nu}^{(1) \mu}-\partial_{\rho} \Gamma_{\lambda \nu}^{(1) \mu} \\ & =\frac{1}{2} \eta^{\mu \sigma}\left(\partial_{\lambda} \partial_{\rho} h_{v \sigma}+\partial_{v} \partial_{\lambda} h_{\rho \sigma}-\partial_{\lambda} \partial_{\sigma} h_{\rho v}-\partial_{\lambda} \partial_{\rho} h_{v \sigma}-\partial_{\nu} \partial_{\rho} h_{\lambda \sigma}+\partial_{\rho} \partial_{\sigma} h_{\lambda \nu}\right) \\ & =\frac{1}{2}\left(\partial_{v} \partial_{\lambda} h_{\rho}^{\mu}+\partial^{\mu} \partial_{\rho} h_{\lambda \nu}-\partial_{v} \partial_{\rho} h_{\lambda}^{\mu}-\partial^{\mu} \partial_{\lambda} h_{\rho v}\right) \\ & =\frac{1}{2} \eta^{\mu \sigma}\left(\partial_{\nu} \partial_{\lambda} h_{\rho \sigma}+\partial_{\sigma} \partial_{\rho} h_{v \lambda}-\partial_{\nu} \partial_{\rho} h_{\lambda \sigma}-\partial_{\sigma} \partial_{\lambda} h_{v \rho}\right) \equiv R_{\nu \lambda \rho}^{(1)^{\mu}} \tag{3.1805} \end{align*}(3.1803)gμνημν+hμv,Γμνλ=12gλρ(μgνρ+νgμρρgμν)(3.1804)12ηλρ(μhvρ+vhμρρhμν)Γμν(1)λ,Rνλρμ=λΓρvμρΓλνμ+ΓλσμΓρvσΓρσμΓλνσλΓρν(1)μρΓλν(1)μ=12ημσ(λρhvσ+vλhρσλσhρvλρhvσνρhλσ+ρσhλν)=12(vλhρμ+μρhλνvρhλμμλhρv)(3.1805)=12ημσ(νλhρσ+σρhvλνρhλσσλhvρ)Rνλρ(1)μ
Choosing μ = i , v = 0 , λ = j μ = i , v = 0 , λ = j mu=i,v=0,lambda=j\mu=i, v=0, \lambda=jμ=i,v=0,λ=j, and ρ = 0 ρ = 0 rho=0\rho=0ρ=0, we have
R 0 j 0 i R ( 1 ) i 0 j 0 = 1 2 η i σ ( 0 j h 0 σ + σ 0 h 0 j 0 0 h j σ σ j h 00 ) (3.1806) = 1 2 ( j 0 h 0 i i 0 h 0 j + 0 0 h j i + i j h 00 ) . R 0 j 0 i R ( 1 ) i 0 j 0 = 1 2 η i σ 0 j h 0 σ + σ 0 h 0 j 0 0 h j σ σ j h 00 (3.1806) = 1 2 j 0 h 0 i i 0 h 0 j + 0 0 h j i + i j h 00 . {:[R_(0j0)^(i)≃R^((1)^(i))_(0j0)=(1)/(2)eta^(i sigma)(del_(0)del_(j)h_(0sigma)+del_(sigma)del_(0)h_(0j)-del_(0)del_(0)h_(j sigma)-del_(sigma)del_(j)h_(00))],[(3.1806)=(1)/(2)(-del_(j)del_(0)h_(0i)-del_(i)del_(0)h_(0j)+del_(0)del_(0)h_(ji)+del_(i)del_(j)h_(00)).]:}\begin{align*} R_{0 j 0}^{i} & \simeq R^{(1)^{i}}{ }_{0 j 0}=\frac{1}{2} \eta^{i \sigma}\left(\partial_{0} \partial_{j} h_{0 \sigma}+\partial_{\sigma} \partial_{0} h_{0 j}-\partial_{0} \partial_{0} h_{j \sigma}-\partial_{\sigma} \partial_{j} h_{00}\right) \\ & =\frac{1}{2}\left(-\partial_{j} \partial_{0} h_{0 i}-\partial_{i} \partial_{0} h_{0 j}+\partial_{0} \partial_{0} h_{j i}+\partial_{i} \partial_{j} h_{00}\right) . \tag{3.1806} \end{align*}R0j0iR(1)i0j0=12ηiσ(0jh0σ+σ0h0j00hjσσjh00)(3.1806)=12(j0h0ii0h0j+00hji+ijh00).
Furthermore, in the Newtonian limit, we have the static field condition, i.e., 0 g α β = 0 0 g α β = 0 del_(0)g_(alpha beta)=0\partial_{0} g_{\alpha \beta}=00gαβ=0, which means that all time derivatives of the metric can be neglected, and thus, we find that
(3.1807) R 0 j 0 i 1 2 i j h 00 (3.1807) R 0 j 0 i 1 2 i j h 00 {:(3.1807)R_(0j0)^(i)≃(1)/(2)del_(i)del_(j)h_(00):}\begin{equation*} R_{0 j 0}^{i} \simeq \frac{1}{2} \partial_{i} \partial_{j} h_{00} \tag{3.1807} \end{equation*}(3.1807)R0j0i12ijh00
Now, h 00 = 2 Φ / c 2 h 00 = 2 Φ / c 2 h_(00)=2Phi//c^(2)h_{00}=2 \Phi / c^{2}h00=2Φ/c2 (see Problem 2.112), so we finally obtain
(3.1808) R 0 j 0 i 1 c 2 i j Φ = 1 c 2 2 Φ x j x i (3.1808) R 0 j 0 i 1 c 2 i j Φ = 1 c 2 2 Φ x j x i {:(3.1808)R_(0j0)^(i)≃(1)/(c^(2))del_(i)del_(j)Phi=(1)/(c^(2))*(del^(2)Phi)/(delx^(j)delx^(i)):}\begin{equation*} R_{0 j 0}^{i} \simeq \frac{1}{c^{2}} \partial_{i} \partial_{j} \Phi=\frac{1}{c^{2}} \cdot \frac{\partial^{2} \Phi}{\partial x^{j} \partial x^{i}} \tag{3.1808} \end{equation*}(3.1808)R0j0i1c2ijΦ=1c22Φxjxi
which, using the result found in a) with τ = t τ = t tau=t\tau=tτ=t, setting c = 1 c = 1 c=1c=1c=1, and again choosing μ = i μ = i mu=i\mu=iμ=i, leads to the equation for the tidal acceleration in Newton's theory of gravitation, namely
(3.1809) d 2 s i d t 2 = 2 Φ x i x j s j (3.1809) d 2 s i d t 2 = 2 Φ x i x j s j {:(3.1809)(d^(2)s^(i))/(dt^(2))=-(del^(2)Phi)/(delx^(i)delx^(j))s^(j):}\begin{equation*} \frac{d^{2} s^{i}}{d t^{2}}=-\frac{\partial^{2} \Phi}{\partial x^{i} \partial x^{j}} s^{j} \tag{3.1809} \end{equation*}(3.1809)d2sidt2=2Φxixjsj
which is what we wanted to show. In words, the acceleration of the separation between the trajectories of the two particles is given by a tensor of the second derivatives of the gravitational potential.

2.115

a) The metric in Minkowski space in Cartesian spatial coordinates is given by
(3.1810) d s 2 = c 2 d t 2 d x 0 2 d y 0 2 d z 0 2 (3.1810) d s 2 = c 2 d t 2 d x 0 2 d y 0 2 d z 0 2 {:(3.1810)ds^(2)=c^(2)dt^(2)-dx_(0)^(2)-dy_(0)^(2)-dz_(0)^(2):}\begin{equation*} d s^{2}=c^{2} d t^{2}-d x_{0}^{2}-d y_{0}^{2}-d z_{0}^{2} \tag{3.1810} \end{equation*}(3.1810)ds2=c2dt2dx02dy02dz02
Consider the motion of a free particle in the lab frame described by a cylindrical coordinate system x = r cos φ , y = r sin φ x = r cos φ , y = r sin φ x=r cos varphi,y=r sin varphix=r \cos \varphi, y=r \sin \varphix=rcosφ,y=rsinφ, and z = z z = z z=zz=zz=z, but also in the rest frame of the particle that is rotating with constant angular velocity ω ω omega\omegaω around the z z zzz-axis relative to the lab frame. The spatial cylindrical coordinates in the rest frame of the particle are given by x = r 0 cos φ 0 , y = r 0 sin φ 0 x = r 0 cos φ 0 , y = r 0 sin φ 0 x=r_(0)cos varphi_(0),y=r_(0)sin varphi_(0)x=r_{0} \cos \varphi_{0}, y=r_{0} \sin \varphi_{0}x=r0cosφ0,y=r0sinφ0, and z = z 0 z = z 0 z=z_(0)z=z_{0}z=z0 and the two frames are related to each other by the following coordinate transformations
(3.1811) r 0 = r , φ 0 = φ ω t , z 0 = z (3.1811) r 0 = r , φ 0 = φ ω t , z 0 = z {:(3.1811)r_(0)=r","quadvarphi_(0)=varphi-omega t","quadz_(0)=z:}\begin{equation*} r_{0}=r, \quad \varphi_{0}=\varphi-\omega t, \quad z_{0}=z \tag{3.1811} \end{equation*}(3.1811)r0=r,φ0=φωt,z0=z
Without loss of generality, we can set c = 1 c = 1 c=1c=1c=1. In the rest frame of the particle, the metric becomes
d s 2 = d t 2 d r 0 2 r 0 2 d φ 0 2 d z 0 2 = d t 2 d r 2 r 2 ( d φ ω d t ) 2 d z 2 (3.1812) = ( 1 ω 2 ) d t 2 d r 2 r 2 d φ 2 d z 2 + 2 r 2 ω d t d φ d s 2 = d t 2 d r 0 2 r 0 2 d φ 0 2 d z 0 2 = d t 2 d r 2 r 2 ( d φ ω d t ) 2 d z 2 (3.1812) = 1 ω 2 d t 2 d r 2 r 2 d φ 2 d z 2 + 2 r 2 ω d t d φ {:[ds^(2)=dt^(2)-dr_(0)^(2)-r_(0)^(2)dvarphi_(0)^(2)-dz_(0)^(2)=dt^(2)-dr^(2)-r^(2)(d varphi-omega dt)^(2)-dz^(2)],[(3.1812)=(1-omega^(2))dt^(2)-dr^(2)-r^(2)dvarphi^(2)-dz^(2)+2r^(2)omega dtd varphi]:}\begin{align*} d s^{2} & =d t^{2}-d r_{0}^{2}-r_{0}^{2} d \varphi_{0}^{2}-d z_{0}^{2}=d t^{2}-d r^{2}-r^{2}(d \varphi-\omega d t)^{2}-d z^{2} \\ & =\left(1-\omega^{2}\right) d t^{2}-d r^{2}-r^{2} d \varphi^{2}-d z^{2}+2 r^{2} \omega d t d \varphi \tag{3.1812} \end{align*}ds2=dt2dr02r02dφ02dz02=dt2dr2r2(dφωdt)2dz2(3.1812)=(1ω2)dt2dr2r2dφ2dz2+2r2ωdtdφ
which in turn implies the Lagrangian
(3.1813) L = g μ ν d x μ d x ν = ( 1 ω 2 r 2 ) t ˙ 2 + 2 ω r 2 t ˙ φ ˙ r ˙ 2 r 2 φ ˙ 2 z ˙ 2 (3.1813) L = g μ ν d x μ d x ν = 1 ω 2 r 2 t ˙ 2 + 2 ω r 2 t ˙ φ ˙ r ˙ 2 r 2 φ ˙ 2 z ˙ 2 {:(3.1813)L=g_(mu nu)dx^(mu)dx^(nu)=(1-omega^(2)r^(2))t^(˙)^(2)+2omegar^(2)t^(˙)varphi^(˙)-r^(˙)^(2)-r^(2)varphi^(˙)^(2)-z^(˙)^(2):}\begin{equation*} \mathcal{L}=g_{\mu \nu} d x^{\mu} d x^{\nu}=\left(1-\omega^{2} r^{2}\right) \dot{t}^{2}+2 \omega r^{2} \dot{t} \dot{\varphi}-\dot{r}^{2}-r^{2} \dot{\varphi}^{2}-\dot{z}^{2} \tag{3.1813} \end{equation*}(3.1813)L=gμνdxμdxν=(1ω2r2)t˙2+2ωr2t˙φ˙r˙2r2φ˙2z˙2
Thus, we can immediately read off the nonzero components of the metric, namely
(3.1814) g t t = 1 ω 2 r 2 , g t φ = g φ t = ω r 2 , g r r = 1 , g φ φ = r 2 , g z z = 1 (3.1814) g t t = 1 ω 2 r 2 , g t φ = g φ t = ω r 2 , g r r = 1 , g φ φ = r 2 , g z z = 1 {:(3.1814)g_(tt)=1-omega^(2)r^(2)","quadg_(t varphi)=g_(varphi t)=omegar^(2)","quadg_(rr)=-1","quadg_(varphi varphi)=-r^(2)","quadg_(zz)=-1:}\begin{equation*} g_{t t}=1-\omega^{2} r^{2}, \quad g_{t \varphi}=g_{\varphi t}=\omega r^{2}, \quad g_{r r}=-1, \quad g_{\varphi \varphi}=-r^{2}, \quad g_{z z}=-1 \tag{3.1814} \end{equation*}(3.1814)gtt=1ω2r2,gtφ=gφt=ωr2,grr=1,gφφ=r2,gzz=1
Now, using the Euler-Lagrange equations, i.e., d d τ L x ˙ μ L x μ = 0 d d τ L x ˙ μ L x μ = 0 (d)/(d tau)(delL)/(delx^(˙)^(mu))-(delL)/(delx^(mu))=0\frac{d}{d \tau} \frac{\partial \mathcal{L}}{\partial \dot{x}^{\mu}}-\frac{\partial \mathcal{L}}{\partial x^{\mu}}=0ddτLx˙μLxμ=0, we find that
L t = 0 , L t ˙ = 2 ( 1 ω 2 r 2 ) t ˙ + 2 ω r 2 φ ˙ (3.1815) t ¨ ω r 1 ω 2 r 2 ( 2 ω t ˙ r ˙ 2 r ˙ φ ˙ r φ ¨ ) = 0 , L r = 2 ω 2 r t ˙ 2 + 4 ω r t ˙ φ ˙ 2 r φ ˙ 2 , L r ˙ = 2 r ˙ (3.1816) r ¨ r φ ˙ 2 ω r ( ω t ˙ 2 22 t ˙ φ ˙ ) = 0 , (3.1817) L φ = 0 , L φ ˙ = 2 ω r 2 t ˙ 2 r 2 φ ˙ φ ¨ + 2 r r ˙ φ ˙ ω ( 2 r t ˙ r ˙ + t ¨ ) = 0 , (3.1818) L z = 0 , L z ˙ = 2 z ˙ d d τ z ˙ = 0 . L t = 0 , L t ˙ = 2 1 ω 2 r 2 t ˙ + 2 ω r 2 φ ˙ (3.1815) t ¨ ω r 1 ω 2 r 2 ( 2 ω t ˙ r ˙ 2 r ˙ φ ˙ r φ ¨ ) = 0 , L r = 2 ω 2 r t ˙ 2 + 4 ω r t ˙ φ ˙ 2 r φ ˙ 2 , L r ˙ = 2 r ˙ (3.1816) r ¨ r φ ˙ 2 ω r ω t ˙ 2 22 t ˙ φ ˙ = 0 , (3.1817) L φ = 0 , L φ ˙ = 2 ω r 2 t ˙ 2 r 2 φ ˙ φ ¨ + 2 r r ˙ φ ˙ ω 2 r t ˙ r ˙ + t ¨ = 0 , (3.1818) L z = 0 , L z ˙ = 2 z ˙ d d τ z ˙ = 0 . {:[(delL)/(del t)=0","quad(delL)/(del(t^(˙)))=2(1-omega^(2)r^(2))t^(˙)+2omegar^(2)varphi^(˙)],[(3.1815)=>quadt^(¨)-(omega r)/(1-omega^(2)r^(2))(2omegat^(˙)r^(˙)-2r^(˙)varphi^(˙)-rvarphi^(¨))=0","],[(delL)/(del r)=-2omega^(2)rt^(˙)^(2)+4omega rt^(˙)varphi^(˙)-2rvarphi^(˙)^(2)","quad(delL)/(del(r^(˙)))=-2r^(˙)],[(3.1816)=>r^(¨)-rvarphi^(˙)^(2)-omega r(omegat^(˙)^(2)-22(t^(˙))(varphi^(˙)))=0","],[(3.1817)(delL)/(del varphi)=0","quad(delL)/(del(varphi^(˙)))=2omegar^(2)t^(˙)-2r^(2)varphi^(˙)=>varphi^(¨)+(2)/(r)r^(˙)varphi^(˙)-omega((2)/(r)(t^(˙))(r^(˙))+(t^(¨)))=0","],[(3.1818)(delL)/(del z)=0","quad(delL)/(del(z^(˙)))=-2z^(˙)=>(d)/(d tau)z^(˙)=0.]:}\begin{align*} \frac{\partial \mathcal{L}}{\partial t} & =0, \quad \frac{\partial \mathcal{L}}{\partial \dot{t}}=2\left(1-\omega^{2} r^{2}\right) \dot{t}+2 \omega r^{2} \dot{\varphi} \\ & \Rightarrow \quad \ddot{t}-\frac{\omega r}{1-\omega^{2} r^{2}}(2 \omega \dot{t} \dot{r}-2 \dot{r} \dot{\varphi}-r \ddot{\varphi})=0, \tag{3.1815}\\ \frac{\partial \mathcal{L}}{\partial r} & =-2 \omega^{2} r \dot{t}^{2}+4 \omega r \dot{t} \dot{\varphi}-2 r \dot{\varphi}^{2}, \quad \frac{\partial \mathcal{L}}{\partial \dot{r}}=-2 \dot{r} \\ & \Rightarrow \ddot{r}-r \dot{\varphi}^{2}-\omega r\left(\omega \dot{t}^{2}-22 \dot{t} \dot{\varphi}\right)=0, \tag{3.1816}\\ \frac{\partial \mathcal{L}}{\partial \varphi} & =0, \quad \frac{\partial \mathcal{L}}{\partial \dot{\varphi}}=2 \omega r^{2} \dot{t}-2 r^{2} \dot{\varphi} \Rightarrow \ddot{\varphi}+\frac{2}{r} \dot{r} \dot{\varphi}-\omega\left(\frac{2}{r} \dot{t} \dot{r}+\ddot{t}\right)=0, \tag{3.1817}\\ \frac{\partial \mathcal{L}}{\partial z} & =0, \quad \frac{\partial \mathcal{L}}{\partial \dot{z}}=-2 \dot{z} \Rightarrow \frac{d}{d \tau} \dot{z}=0 . \tag{3.1818} \end{align*}Lt=0,Lt˙=2(1ω2r2)t˙+2ωr2φ˙(3.1815)t¨ωr1ω2r2(2ωt˙r˙2r˙φ˙rφ¨)=0,Lr=2ω2rt˙2+4ωrt˙φ˙2rφ˙2,Lr˙=2r˙(3.1816)r¨rφ˙2ωr(ωt˙222t˙φ˙)=0,(3.1817)Lφ=0,Lφ˙=2ωr2t˙2r2φ˙φ¨+2rr˙φ˙ω(2rt˙r˙+t¨)=0,(3.1818)Lz=0,Lz˙=2z˙ddτz˙=0.
Since L t = L φ = L z = 0 L t = L φ = L z = 0 (delL)/(del t)=(delL)/(del varphi)=(delL)/(del z)=0\frac{\partial \mathcal{L}}{\partial t}=\frac{\partial \mathcal{L}}{\partial \varphi}=\frac{\partial \mathcal{L}}{\partial z}=0Lt=Lφ=Lz=0, we basically have three constants of motion, i.e., L t = L t = (delL)/(del t)=\frac{\partial \mathcal{L}}{\partial t}=Lt= const., L φ ˙ = L φ ˙ = (delL)/(del(varphi^(˙)))=\frac{\partial \mathcal{L}}{\partial \dot{\varphi}}=Lφ˙= const., and L z = L z = (delL)/(del z)=\frac{\partial \mathcal{L}}{\partial z}=Lz= const. However, we observe that two of the EulerLagrange equations contain both t ¨ t ¨ t^(¨)\ddot{t}t¨ and φ ¨ φ ¨ varphi^(¨)\ddot{\varphi}φ¨, and therefore, solving for t ¨ t ¨ t^(¨)\ddot{t}t¨ and φ ¨ φ ¨ varphi^(¨)\ddot{\varphi}φ¨, we obtain the geodesic equations
(3.1819) t ¨ = 0 , r ¨ ω 2 r t ˙ 2 + 2 ω r t ˙ φ ˙ r φ ˙ 2 = 0 , φ ¨ 2 ω r t ˙ r ˙ + 2 r r ˙ φ ˙ = 0 , z ¨ = 0 (3.1819) t ¨ = 0 , r ¨ ω 2 r t ˙ 2 + 2 ω r t ˙ φ ˙ r φ ˙ 2 = 0 , φ ¨ 2 ω r t ˙ r ˙ + 2 r r ˙ φ ˙ = 0 , z ¨ = 0 {:(3.1819)t^(¨)=0","quadr^(¨)-omega^(2)rt^(˙)^(2)+2omega rt^(˙)varphi^(˙)-rvarphi^(˙)^(2)=0","quadvarphi^(¨)-(2omega)/(r)t^(˙)r^(˙)+(2)/(r)r^(˙)varphi^(˙)=0","quadz^(¨)=0:}\begin{equation*} \ddot{t}=0, \quad \ddot{r}-\omega^{2} r \dot{t}^{2}+2 \omega r \dot{t} \dot{\varphi}-r \dot{\varphi}^{2}=0, \quad \ddot{\varphi}-\frac{2 \omega}{r} \dot{t} \dot{r}+\frac{2}{r} \dot{r} \dot{\varphi}=0, \quad \ddot{z}=0 \tag{3.1819} \end{equation*}(3.1819)t¨=0,r¨ω2rt˙2+2ωrt˙φ˙rφ˙2=0,φ¨2ωrt˙r˙+2rr˙φ˙=0,z¨=0
Thus, comparing the geodesic equations with the general formula for the geodesic equations, i.e., x ¨ μ + Γ ν λ μ x ˙ ν x ˙ λ = 0 x ¨ μ + Γ ν λ μ x ˙ ν x ˙ λ = 0 x^(¨)^(mu)+Gamma_(nu lambda)^(mu)x^(˙)^(nu)x^(˙)^(lambda)=0\ddot{x}^{\mu}+\Gamma_{\nu \lambda}^{\mu} \dot{x}^{\nu} \dot{x}^{\lambda}=0x¨μ+Γνλμx˙νx˙λ=0, and using Γ ν λ μ = Γ λ ν μ Γ ν λ μ = Γ λ ν μ Gamma_(nu lambda)^(mu)=Gamma_(lambda nu)^(mu)\Gamma_{\nu \lambda}^{\mu}=\Gamma_{\lambda \nu}^{\mu}Γνλμ=Γλνμ, we can read off the eight nonzero Christoffel symbols as
Γ t t r = ω 2 r , Γ t φ r = Γ φ t r = ω r , Γ φ φ r = r (3.1820) Γ t r φ = Γ r t φ = ω r , Γ r φ φ = Γ φ r φ = 1 r Γ t t r = ω 2 r , Γ t φ r = Γ φ t r = ω r , Γ φ φ r = r (3.1820) Γ t r φ = Γ r t φ = ω r , Γ r φ φ = Γ φ r φ = 1 r {:[Gamma_(tt)^(r)=-omega^(2)r","quadGamma_(t varphi)^(r)=Gamma_(varphi t)^(r)=omega r","quadGamma_(varphi varphi)^(r)=-r],[(3.1820)Gamma_(tr)^(varphi)=Gamma_(rt)^(varphi)=-(omega )/(r)","quadGamma_(r varphi)^(varphi)=Gamma_(varphi r)^(varphi)=(1)/(r)]:}\begin{gather*} \Gamma_{t t}^{r}=-\omega^{2} r, \quad \Gamma_{t \varphi}^{r}=\Gamma_{\varphi t}^{r}=\omega r, \quad \Gamma_{\varphi \varphi}^{r}=-r \\ \Gamma_{t r}^{\varphi}=\Gamma_{r t}^{\varphi}=-\frac{\omega}{r}, \quad \Gamma_{r \varphi}^{\varphi}=\Gamma_{\varphi r}^{\varphi}=\frac{1}{r} \tag{3.1820} \end{gather*}Γttr=ω2r,Γtφr=Γφtr=ωr,Γφφr=r(3.1820)Γtrφ=Γrtφ=ωr,Γrφφ=Γφrφ=1r
Note that in the case we set ω = 0 ω = 0 omega=0\omega=0ω=0, the Christoffel symbols reduce to three, i.e., Γ φ φ r = r Γ φ φ r = r Gamma_(varphi varphi)^(r)=-r\Gamma_{\varphi \varphi}^{r}=-rΓφφr=r and Γ r φ φ = Γ φ r φ = 1 r Γ r φ φ = Γ φ r φ = 1 r Gamma_(r varphi)^(varphi)=Gamma_(varphi r)^(varphi)=(1)/(r)\Gamma_{r \varphi}^{\varphi}=\Gamma_{\varphi r}^{\varphi}=\frac{1}{r}Γrφφ=Γφrφ=1r, which are the Christoffel symbols in nonrotating cylinder coordinates.
To summarize, in the rest frame of a free particle rotating with constant angular velocity ω ω omega\omegaω around the z z zzz-coordinate axis, the Christoffel symbols and the geodesic equations are
Γ t t r = ω 2 r , Γ t φ r = Γ φ t r = ω r , Γ φ φ r = r Γ t r φ = Γ r t φ = ω r , Γ r φ φ = Γ φ r φ = 1 r (3.1821) t ¨ = 0 , r ¨ ω 2 r t ˙ 2 + 2 ω r t ˙ φ ˙ r φ ˙ 2 = 0 , φ ¨ 2 ω r t r ˙ + 2 r r ˙ φ ˙ = 0 , z ¨ = 0 Γ t t r = ω 2 r , Γ t φ r = Γ φ t r = ω r , Γ φ φ r = r Γ t r φ = Γ r t φ = ω r , Γ r φ φ = Γ φ r φ = 1 r (3.1821) t ¨ = 0 , r ¨ ω 2 r t ˙ 2 + 2 ω r t ˙ φ ˙ r φ ˙ 2 = 0 , φ ¨ 2 ω r t r ˙ + 2 r r ˙ φ ˙ = 0 , z ¨ = 0 {:[Gamma_(tt)^(r)=-omega^(2)r","quadGamma_(t varphi)^(r)=Gamma_(varphi t)^(r)=omega r","quadGamma_(varphi varphi)^(r)=-r],[Gamma_(tr)^(varphi)=Gamma_(rt)^(varphi)=-(omega )/(r)","quadGamma_(r varphi)^(varphi)=Gamma_(varphi r)^(varphi)=(1)/(r)],[(3.1821)t^(¨)=0","quadr^(¨)-omega^(2)rt^(˙)^(2)+2omega rt^(˙)varphi^(˙)-rvarphi^(˙)^(2)=0","quadvarphi^(¨)-(2omega)/(r)tr^(˙)+(2)/(r)r^(˙)varphi^(˙)=0","quadz^(¨)=0]:}\begin{align*} & \Gamma_{t t}^{r}=-\omega^{2} r, \quad \Gamma_{t \varphi}^{r}=\Gamma_{\varphi t}^{r}=\omega r, \quad \Gamma_{\varphi \varphi}^{r}=-r \\ & \Gamma_{t r}^{\varphi}=\Gamma_{r t}^{\varphi}=-\frac{\omega}{r}, \quad \Gamma_{r \varphi}^{\varphi}=\Gamma_{\varphi r}^{\varphi}=\frac{1}{r} \\ & \ddot{t}=0, \quad \ddot{r}-\omega^{2} r \dot{t}^{2}+2 \omega r \dot{t} \dot{\varphi}-r \dot{\varphi}^{2}=0, \quad \ddot{\varphi}-\frac{2 \omega}{r} t \dot{r}+\frac{2}{r} \dot{r} \dot{\varphi}=0, \quad \ddot{z}=0 \tag{3.1821} \end{align*}Γttr=ω2r,Γtφr=Γφtr=ωr,Γφφr=rΓtrφ=Γrtφ=ωr,Γrφφ=Γφrφ=1r(3.1821)t¨=0,r¨ω2rt˙2+2ωrt˙φ˙rφ˙2=0,φ¨2ωrtr˙+2rr˙φ˙=0,z¨=0
b) Now, in the Newtonian limit, the approximate geodesic equations (see also Problem 2.111) are given by
(3.1822) d 2 x μ d σ 2 + Γ 00 μ ( d x 0 d σ ) 2 = 0 (3.1822) d 2 x μ d σ 2 + Γ 00 μ d x 0 d σ 2 = 0 {:(3.1822)(d^(2)x^(mu))/(dsigma^(2))+Gamma_(00)^(mu)((dx^(0))/(d sigma))^(2)=0:}\begin{equation*} \frac{d^{2} x^{\mu}}{d \sigma^{2}}+\Gamma_{00}^{\mu}\left(\frac{d x^{0}}{d \sigma}\right)^{2}=0 \tag{3.1822} \end{equation*}(3.1822)d2xμdσ2+Γ00μ(dx0dσ)2=0
since for small velocities, the component x ˙ 0 ( σ ) x ˙ 0 ( σ ) x^(˙)^(0)(sigma)\dot{x}^{0}(\sigma)x˙0(σ) of the 4 -velocity is much larger than the corresponding spatial components, and where σ σ sigma\sigmaσ is the curve parameter. Using the results found in a), we only have one nonzero Christoffel symbol on the form Γ 00 μ Γ 00 μ Gamma_(00)^(mu)\Gamma_{00}^{\mu}Γ00μ, which is Γ 00 r = Γ t t r = ω 2 r Γ 00 r = Γ t t r = ω 2 r Gamma_(00)^(r)=Gamma_(tt)^(r)=-omega^(2)r\Gamma_{00}^{r}=\Gamma_{t t}^{r}=-\omega^{2} rΓ00r=Γttr=ω2r, and therefore, the approximate geodesic equations are reduced to
(3.1823) d 2 t d σ 2 = 0 , d 2 r d σ 2 ω 2 r ( d t d σ ) 2 = 0 (3.1823) d 2 t d σ 2 = 0 , d 2 r d σ 2 ω 2 r d t d σ 2 = 0 {:(3.1823)(d^(2)t)/(dsigma^(2))=0","quad(d^(2)r)/(dsigma^(2))-omega^(2)r((dt)/(d sigma))^(2)=0:}\begin{equation*} \frac{d^{2} t}{d \sigma^{2}}=0, \quad \frac{d^{2} r}{d \sigma^{2}}-\omega^{2} r\left(\frac{d t}{d \sigma}\right)^{2}=0 \tag{3.1823} \end{equation*}(3.1823)d2tdσ2=0,d2rdσ2ω2r(dtdσ)2=0
In the rest frame of the particle, the first equation means that we can choose the time t t ttt as the curve parameter σ σ sigma\sigmaσ, i.e., t = σ t = σ t=sigmat=\sigmat=σ, and then the second equation becomes
(3.1824) r ¨ ω 2 r = 0 (3.1824) r ¨ ω 2 r = 0 {:(3.1824)r^(¨)-omega^(2)r=0:}\begin{equation*} \ddot{r}-\omega^{2} r=0 \tag{3.1824} \end{equation*}(3.1824)r¨ω2r=0
Multiplying this equation with the mass m m mmm of the particle, we obtain
(3.1825) m r ¨ = m ω 2 r F centrifugal , (3.1825) m r ¨ = m ω 2 r F centrifugal  , {:(3.1825)mr^(¨)=momega^(2)r-=F_("centrifugal ")",":}\begin{equation*} m \ddot{r}=m \omega^{2} r \equiv F_{\text {centrifugal }}, \tag{3.1825} \end{equation*}(3.1825)mr¨=mω2rFcentrifugal ,
where F centrifugal = m ω 2 r F centrifugal  = m ω 2 r F_("centrifugal ")=momega^(2)rF_{\text {centrifugal }}=m \omega^{2} rFcentrifugal =mω2r is the centrifugal force that appears to act on all objects when viewed in a rotating frame.
Thus, in the nonrelativistic limit, Newton's equation of motion in the rotating frame of the free particle is given by
(3.1826) m r ¨ = m ω 2 r . (3.1826) m r ¨ = m ω 2 r . {:(3.1826)mr^(¨)=momega^(2)r.:}\begin{equation*} m \ddot{r}=m \omega^{2} r . \tag{3.1826} \end{equation*}(3.1826)mr¨=mω2r.
Note that this equation is expressed in the radial coordinate of either the lab frame or the rest frame of the particle, since r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0 [see a)].

2.116

a) In spherical coordinates the Lagrangian is
(3.1827) L = 1 2 m r ˙ 2 + G M m r = 1 2 m ( r ˙ 2 + r 2 φ ˙ 2 ) + G M m r (3.1827) L = 1 2 m r ˙ 2 + G M m r = 1 2 m r ˙ 2 + r 2 φ ˙ 2 + G M m r {:(3.1827)L=(1)/(2)mr^(˙)^(2)+(GMm)/(r)=(1)/(2)m(r^(˙)^(2)+r^(2)varphi^(˙)^(2))+(GMm)/(r):}\begin{equation*} \mathcal{L}=\frac{1}{2} m \dot{\mathbf{r}}^{2}+\frac{G M m}{r}=\frac{1}{2} m\left(\dot{r}^{2}+r^{2} \dot{\varphi}^{2}\right)+\frac{G M m}{r} \tag{3.1827} \end{equation*}(3.1827)L=12mr˙2+GMmr=12m(r˙2+r2φ˙2)+GMmr
where it is assumed that θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2 and θ ˙ = 0 θ ˙ = 0 theta^(˙)=0\dot{\theta}=0θ˙=0 with dot meaning differentiation with respect to time t t ttt. This yields the Euler-Lagrange equations
(3.1828) d d t L r ˙ L r = m r φ ˙ 2 + G M m r 2 = 0 (3.1829) d d t L φ ˙ L φ = d d t ( m r 2 φ ˙ ) = 0 (3.1828) d d t L r ˙ L r = m r φ ˙ 2 + G M m r 2 = 0 (3.1829) d d t L φ ˙ L φ = d d t m r 2 φ ˙ = 0 {:[(3.1828)(d)/(dt)(delL)/(del(r^(˙)))-(delL)/(del r)=-mrvarphi^(˙)^(2)+(GMm)/(r^(2))=0],[(3.1829)(d)/(dt)*(delL)/(del(varphi^(˙)))-(delL)/(del varphi)=(d)/(dt)(mr^(2)(varphi^(˙)))=0]:}\begin{align*} & \frac{d}{d t} \frac{\partial \mathcal{L}}{\partial \dot{r}}-\frac{\partial \mathcal{L}}{\partial r}=-m r \dot{\varphi}^{2}+\frac{G M m}{r^{2}}=0 \tag{3.1828}\\ & \frac{d}{d t} \cdot \frac{\partial \mathcal{L}}{\partial \dot{\varphi}}-\frac{\partial \mathcal{L}}{\partial \varphi}=\frac{d}{d t}\left(m r^{2} \dot{\varphi}\right)=0 \tag{3.1829} \end{align*}(3.1828)ddtLr˙Lr=mrφ˙2+GMmr2=0(3.1829)ddtLφ˙Lφ=ddt(mr2φ˙)=0
where it holds that r ˙ = r ¨ = 0 r ˙ = r ¨ = 0 r^(˙)=r^(¨)=0\dot{r}=\ddot{r}=0r˙=r¨=0, since it is assumed that r = r 0 = r = r 0 = r=r_(0)=r=r_{0}=r=r0= const. The second equation is solved by L = m r 2 φ ˙ = L = m r 2 φ ˙ = L=mr^(2)varphi^(˙)=L=m r^{2} \dot{\varphi}=L=mr2φ˙= constant, i.e., φ ˙ = L / ( m r 2 ) φ ˙ = L / m r 2 varphi^(˙)=L//(mr^(2))\dot{\varphi}=L /\left(m r^{2}\right)φ˙=L/(mr2). Inserting this into the first equation, we obtain
(3.1830) m r ( L m r 2 ) 2 + G M m r 2 = 0 r = L 2 G M m 2 r 0 (3.1830) m r L m r 2 2 + G M m r 2 = 0 r = L 2 G M m 2 r 0 {:(3.1830)-mr((L)/(mr^(2)))^(2)+(GMm)/(r^(2))=0=>r=(L^(2))/(GMm^(2))-=r_(0):}\begin{equation*} -m r\left(\frac{L}{m r^{2}}\right)^{2}+\frac{G M m}{r^{2}}=0 \Rightarrow r=\frac{L^{2}}{G M m^{2}} \equiv r_{0} \tag{3.1830} \end{equation*}(3.1830)mr(Lmr2)2+GMmr2=0r=L2GMm2r0
which is the trajectory of the planet.
Note that r 0 r 0 r_(0)r_{0}r0 is exactly the radius where the centrifugal force m r 2 φ ˙ m r 2 φ ˙ mr^(2)varphi^(˙)m r^{2} \dot{\varphi}mr2φ˙ is balanced by the attractive gravitational force G M m / r 2 G M m / r 2 GMm//r^(2)G M m / r^{2}GMm/r2, i.e., m r 2 φ ˙ = G M m / r 2 m r 2 φ ˙ = G M m / r 2 mr^(2)varphi^(˙)=GMm//r^(2)m r^{2} \dot{\varphi}=G M m / r^{2}mr2φ˙=GMm/r2, which together with L = m r 2 φ ˙ L = m r 2 φ ˙ L=mr^(2)varphi^(˙)L=m r^{2} \dot{\varphi}L=mr2φ˙ gives an elementary derivation of the trajectory of the planet.
b) The result in a) can be summarized as follows: It has been shown that m r ¨ = m Φ eff ( r ) m r ¨ = m Φ eff  ( r ) mr^(¨)=-mPhi_("eff ")^(')(r)m \ddot{r}=-m \Phi_{\text {eff }}^{\prime}(r)mr¨=mΦeff (r) with the effective gravitational potential
(3.1831) m Φ eff ( r ) = G M m r + 1 2 m r 2 φ ˙ 2 = m ( G M r + L 2 2 m 2 r 2 ) (3.1831) m Φ eff ( r ) = G M m r + 1 2 m r 2 φ ˙ 2 = m G M r + L 2 2 m 2 r 2 {:(3.1831)mPhi_(eff)(r)=-(GMm)/(r)+(1)/(2)mr^(2)varphi^(˙)^(2)=m(-(GM)/(r)+(L^(2))/(2m^(2)r^(2))):}\begin{equation*} m \Phi_{\mathrm{eff}}(r)=-\frac{G M m}{r}+\frac{1}{2} m r^{2} \dot{\varphi}^{2}=m\left(-\frac{G M}{r}+\frac{L^{2}}{2 m^{2} r^{2}}\right) \tag{3.1831} \end{equation*}(3.1831)mΦeff(r)=GMmr+12mr2φ˙2=m(GMr+L22m2r2)
The stationary orbit r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0 can thus be obtained as the solution to Φ eff ( r ) = 0 Φ eff  ( r ) = 0 Phi_("eff ")^(')(r)=0\Phi_{\text {eff }}^{\prime}(r)=0Φeff (r)=0. This solution r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0 is a minimum of Φ eff ( r ) Φ eff  ( r ) Phi_("eff ")(r)\Phi_{\text {eff }}(r)Φeff (r), and therefore, it corresponds to a stable orbit.
There exists a natural generalization of the effective potential Φ eff ( r ) Φ eff  ( r ) Phi_("eff ")(r)\Phi_{\text {eff }}(r)Φeff (r) to general relativity, namely
(3.1832) Φ ( r ) = r 2 r + L 2 2 m 2 r 2 r L 2 2 m 2 r 3 , (3.1832) Φ ( r ) = r 2 r + L 2 2 m 2 r 2 r L 2 2 m 2 r 3 , {:(3.1832)Phi(r)=-(r_(**))/(2r)+(L^(2))/(2m^(2)r^(2))-(r_(**)L^(2))/(2m^(2)r^(3))",":}\begin{equation*} \Phi(r)=-\frac{r_{*}}{2 r}+\frac{L^{2}}{2 m^{2} r^{2}}-\frac{r_{*} L^{2}}{2 m^{2} r^{3}}, \tag{3.1832} \end{equation*}(3.1832)Φ(r)=r2r+L22m2r2rL22m2r3,
where r = 2 G M r = 2 G M r_(**)=2GMr_{*}=2 G Mr=2GM is the Schwarzschild radius, which can be written as
(3.1833) Φ ( r ) = r 2 r + r 0 r 4 r 2 r 0 r 2 4 r 3 (3.1833) Φ ( r ) = r 2 r + r 0 r 4 r 2 r 0 r 2 4 r 3 {:(3.1833)Phi(r)=-(r_(**))/(2r)+(r_(0)r_(**))/(4r^(2))-(r_(0)r_(**)^(2))/(4r^(3)):}\begin{equation*} \Phi(r)=-\frac{r_{*}}{2 r}+\frac{r_{0} r_{*}}{4 r^{2}}-\frac{r_{0} r_{*}^{2}}{4 r^{3}} \tag{3.1833} \end{equation*}(3.1833)Φ(r)=r2r+r0r4r2r0r24r3
where r 0 L 2 / ( G M m 2 ) r 0 L 2 / G M m 2 r_(0)-=L^(2)//(GMm^(2))r_{0} \equiv L^{2} /\left(G M m^{2}\right)r0L2/(GMm2), which is the solution to the problem in a). The general relativity generalization of the orbit r 0 r 0 r_(0)r_{0}r0 can be computed by solving Φ ( r ) = 0 Φ ( r ) = 0 Phi^(')(r)=0\Phi^{\prime}(r)=0Φ(r)=0, which yields
(3.1834) r 2 r 0 r + 3 2 r 0 r = 0 (3.1834) r 2 r 0 r + 3 2 r 0 r = 0 {:(3.1834)r^(2)-r_(0)r+(3)/(2)r_(0)r_(**)=0:}\begin{equation*} r^{2}-r_{0} r+\frac{3}{2} r_{0} r_{*}=0 \tag{3.1834} \end{equation*}(3.1834)r2r0r+32r0r=0
Thus, we have two possible solutions
(3.1835) r ± = r 0 2 ( 1 ± 1 6 r r 0 ) (3.1835) r ± = r 0 2 1 ± 1 6 r r 0 {:(3.1835)r_(+-)=(r_(0))/(2)(1+-sqrt(1-(6r_(**))/(r_(0)))):}\begin{equation*} r_{ \pm}=\frac{r_{0}}{2}\left(1 \pm \sqrt{1-\frac{6 r_{*}}{r_{0}}}\right) \tag{3.1835} \end{equation*}(3.1835)r±=r02(1±16rr0)
The solution r + r + r_(+)r_{+}r+corresponds to a minimum of Φ eff ( r ) Φ eff  ( r ) Phi_("eff ")(r)\Phi_{\text {eff }}(r)Φeff (r) and is thus a stable orbit generalizing the Newtonian solution found in a), whereas the solution r r r_(-)r_{-}rcorresponds to a maximum and is thus unstable, but r 0 r 0 r_(-)rarr0r_{-} \rightarrow 0r0 in the Newtonian limit r 0 r 0 r_(**)rarr0r_{*} \rightarrow 0r0. Therefore, the generalization of the orbit r 0 r 0 r_(0)r_{0}r0 in a) to general relativity is
(3.1836) r + = r 0 2 ( 1 + 1 6 r r 0 ) . (3.1836) r + = r 0 2 1 + 1 6 r r 0 . {:(3.1836)r_(+)=(r_(0))/(2)(1+sqrt(1-(6r_(**))/(r_(0)))).:}\begin{equation*} r_{+}=\frac{r_{0}}{2}\left(1+\sqrt{1-\frac{6 r_{*}}{r_{0}}}\right) . \tag{3.1836} \end{equation*}(3.1836)r+=r02(1+16rr0).

2.117

a) We need to assume the weak-field limit where the metric can be written
(3.1837) g μ ν = η μ ν + h μ ν (3.1837) g μ ν = η μ ν + h μ ν {:(3.1837)g_(mu nu)=eta_(mu nu)+h_(mu nu):}\begin{equation*} g_{\mu \nu}=\eta_{\mu \nu}+h_{\mu \nu} \tag{3.1837} \end{equation*}(3.1837)gμν=ημν+hμν
where η μ ν η μ ν eta_(mu nu)\eta_{\mu \nu}ημν is the Minkowski metric and h μ ν h μ ν h_(mu nu)h_{\mu \nu}hμν is a small perturbation. We will only need to determine the Riemann tensor to first order in h μ ν h μ ν h_(mu nu)h_{\mu \nu}hμν. Furthermore, we assume that the source of the gravitational field is not large and slowly moving, hence only the component T 00 = c 2 ρ T 00 = c 2 ρ T^(00)=c^(2)rhoT^{00}=c^{2} \rhoT00=c2ρ gives a relevant contribution to the field. Furthermore, we assume that in the Newtonian limit ρ p c 2 ρ p c 2 rho≫(p)/(c^(2))\rho \gg \frac{p}{c^{2}}ρpc2 hence we can assume T μ ν = ρ U μ U ν T μ ν = ρ U μ U ν T^(mu nu)=rhoU^(mu)U^(nu)T^{\mu \nu}=\rho U^{\mu} U^{\nu}Tμν=ρUμUν.
b) Einstein's equations are
(3.1838) G μ ν = 8 π G c 4 T μ ν (3.1838) G μ ν = 8 π G c 4 T μ ν {:(3.1838)G^(mu nu)=8pi(G)/(c^(4))T^(mu nu):}\begin{equation*} G^{\mu \nu}=8 \pi \frac{G}{c^{4}} T^{\mu \nu} \tag{3.1838} \end{equation*}(3.1838)Gμν=8πGc4Tμν
where G μ ν = R μ ν 1 2 R g μ ν G μ ν = R μ ν 1 2 R g μ ν G^(mu nu)=R^(mu nu)-(1)/(2)Rg^(mu nu)G^{\mu \nu}=R^{\mu \nu}-\frac{1}{2} R g^{\mu \nu}Gμν=Rμν12Rgμν. For the metric in g μ ν = η μ ν + h μ ν g μ ν = η μ ν + h μ ν g_(mu nu)=eta_(mu nu)+h_(mu nu)g_{\mu \nu}=\eta_{\mu \nu}+h_{\mu \nu}gμν=ημν+hμν the Christoffel symbols are given by
(3.1839) Γ μ ν ω = 1 2 η ω ρ ( μ h v ρ + ν h μ ρ ρ h μ v ) (3.1839) Γ μ ν ω = 1 2 η ω ρ μ h v ρ + ν h μ ρ ρ h μ v {:(3.1839)Gamma_(mu nu)^(omega)=(1)/(2)eta^(omega rho)(del_(mu)h_(v rho)+del_(nu)h_(mu rho)-del_(rho)h_(mu v)):}\begin{equation*} \Gamma_{\mu \nu}^{\omega}=\frac{1}{2} \eta^{\omega \rho}\left(\partial_{\mu} h_{v \rho}+\partial_{\nu} h_{\mu \rho}-\partial_{\rho} h_{\mu v}\right) \tag{3.1839} \end{equation*}(3.1839)Γμνω=12ηωρ(μhvρ+νhμρρhμv)
Since the lowest order of h μ ν h μ ν h_(mu nu)h_{\mu \nu}hμν in the Christoffel symbols is the first, the Riemann tensor of first order will therefore only contain the derivative terms
(3.1840) R ω μ ν λ = 1 2 ( μ Γ ν ω λ ν Γ μ ω λ ) = 1 2 ( ω μ h v λ + λ ν h ω μ ω ν h μ λ λ μ h ω v ) . (3.1840) R ω μ ν λ = 1 2 μ Γ ν ω λ ν Γ μ ω λ = 1 2 ω μ h v λ + λ ν h ω μ ω ν h μ λ λ μ h ω v . {:(3.1840)R_(omega mu nu)^(lambda)=(1)/(2)(del_(mu)Gamma_(nu omega)^(lambda)-del_(nu)Gamma_(mu omega)^(lambda))=(1)/(2)(del_(omega)del_(mu)h_(v)^(lambda)+del^(lambda)del_(nu)h_(omega mu)-del_(omega)del_(nu)h_(mu)^(lambda)-del^(lambda)del_(mu)h_(omega v)).:}\begin{equation*} R_{\omega \mu \nu}^{\lambda}=\frac{1}{2}\left(\partial_{\mu} \Gamma_{\nu \omega}^{\lambda}-\partial_{\nu} \Gamma_{\mu \omega}^{\lambda}\right)=\frac{1}{2}\left(\partial_{\omega} \partial_{\mu} h_{v}^{\lambda}+\partial^{\lambda} \partial_{\nu} h_{\omega \mu}-\partial_{\omega} \partial_{\nu} h_{\mu}^{\lambda}-\partial^{\lambda} \partial_{\mu} h_{\omega v}\right) . \tag{3.1840} \end{equation*}(3.1840)Rωμνλ=12(μΓνωλνΓμωλ)=12(ωμhvλ+λνhωμωνhμλλμhωv).
Using the gauge condition λ h μ λ = μ h / 2 λ h μ λ = μ h / 2 del^(lambda)h_(mu lambda)=del_(mu)h//2\partial^{\lambda} h_{\mu \lambda}=\partial_{\mu} h / 2λhμλ=μh/2, we can write the Ricci tensor and Ricci scalar as
(3.1841) R μ ν = 1 2 h μ ν , R = 1 2 h , = μ μ . (3.1841) R μ ν = 1 2 h μ ν , R = 1 2 h , = μ μ . {:(3.1841)R_(mu nu)=-(1)/(2)◻h_(mu nu)","quad R=(1)/(2)◻h","quad◻=del_(mu)del^(mu).:}\begin{equation*} R_{\mu \nu}=-\frac{1}{2} \square h_{\mu \nu}, \quad R=\frac{1}{2} \square h, \quad \square=\partial_{\mu} \partial^{\mu} . \tag{3.1841} \end{equation*}(3.1841)Rμν=12hμν,R=12h,=μμ.
Einstein's equations can now be written as
(3.1842) 1 2 ( h μ ν 1 2 h η μ ν ) = 8 π G c 4 T μ ν . (3.1842) 1 2 h μ ν 1 2 h η μ ν = 8 π G c 4 T μ ν . {:(3.1842)-(1)/(2)◻(h^(mu nu)-(1)/(2)heta^(mu nu))=8pi(G)/(c^(4))T^(mu nu).:}\begin{equation*} -\frac{1}{2} \square\left(h^{\mu \nu}-\frac{1}{2} h \eta^{\mu \nu}\right)=8 \pi \frac{G}{c^{4}} T^{\mu \nu} . \tag{3.1842} \end{equation*}(3.1842)12(hμν12hημν)=8πGc4Tμν.
Since only the 00-component gives a sizable contribution, we obtain
(3.1843) ( h 00 1 2 h ) = 16 π G c 2 ρ (3.1843) h 00 1 2 h = 16 π G c 2 ρ {:(3.1843)◻(h^(00)-(1)/(2)h)=-16 pi(G)/(c^(2))rho:}\begin{equation*} \square\left(h^{00}-\frac{1}{2} h\right)=-16 \pi \frac{G}{c^{2}} \rho \tag{3.1843} \end{equation*}(3.1843)(h0012h)=16πGc2ρ
Since the source was moving slowly we can drop the 0-component of the derivative 0 = 1 c t 0 = 1 c t del_(0)=(1)/(c)del_(t)\partial_{0}=\frac{1}{c} \partial_{t}0=1ct. Thus,
(3.1844) 2 ( h 00 1 2 h ) = 16 π G c 2 ρ (3.1844) 2 h 00 1 2 h = 16 π G c 2 ρ {:(3.1844)grad^(2)(h^(00)-(1)/(2)h)=16 pi(G)/(c^(2))*rho:}\begin{equation*} \nabla^{2}\left(h^{00}-\frac{1}{2} h\right)=16 \pi \frac{G}{c^{2}} \cdot \rho \tag{3.1844} \end{equation*}(3.1844)2(h0012h)=16πGc2ρ
Hence, with
(3.1845) h 00 1 2 h = 4 c 2 ϕ (3.1845) h 00 1 2 h = 4 c 2 ϕ {:(3.1845)h^(00)-(1)/(2)h=(4)/(c^(2))phi:}\begin{equation*} h^{00}-\frac{1}{2} h=\frac{4}{c^{2}} \phi \tag{3.1845} \end{equation*}(3.1845)h0012h=4c2ϕ
we find 2 ϕ = 4 π G ρ 2 ϕ = 4 π G ρ grad^(2)phi=4pi G rho\nabla^{2} \phi=4 \pi G \rho2ϕ=4πGρ.

2.118

The metric outside of the Earth is given by
(3.1846) d s 2 = ( 1 + 2 Φ ) d t 2 + ( 1 + 2 Φ ) 1 d r 2 r 2 d Ω 2 (3.1846) d s 2 = ( 1 + 2 Φ ) d t 2 + ( 1 + 2 Φ ) 1 d r 2 r 2 d Ω 2 {:(3.1846)ds^(2)=(1+2Phi)dt^(2)+(1+2Phi)^(-1)dr^(2)-r^(2)dOmega^(2):}\begin{equation*} d s^{2}=(1+2 \Phi) d t^{2}+(1+2 \Phi)^{-1} d r^{2}-r^{2} d \Omega^{2} \tag{3.1846} \end{equation*}(3.1846)ds2=(1+2Φ)dt2+(1+2Φ)1dr2r2dΩ2
where Φ = G M r Φ = G M r Phi=-(GM)/(r)\Phi=-\frac{G M}{r}Φ=GMr is the gravitational potential and d Ω 2 = d θ 2 + sin 2 θ d ϕ 2 d Ω 2 = d θ 2 + sin 2 θ d ϕ 2 dOmega^(2)=dtheta^(2)+sin^(2)theta dphi^(2)d \Omega^{2}=d \theta^{2}+\sin ^{2} \theta d \phi^{2}dΩ2=dθ2+sin2θdϕ2. A satellite is orbiting at a distance R 1 R 1 R_(1)R_{1}R1 from the surface. We are interested the eigentime for the satellite to complete a full orbit around the Earth. Therefore, we have R = R E + R 1 R = R E + R 1 R=R_(E)+R_(1)R=R_{E}+R_{1}R=RE+R1 and due to the spherical symmetry we can assume θ = π 2 θ = π 2 theta=(pi)/(2)\theta=\frac{\pi}{2}θ=π2. Thus, the metric can be written as
(3.1847) d s 2 = ( 1 + 2 Φ ) d t 2 ( 1 + 2 Φ ) 1 d r 2 R 2 d ϕ 2 (3.1847) d s 2 = ( 1 + 2 Φ ) d t 2 ( 1 + 2 Φ ) 1 d r 2 R 2 d ϕ 2 {:(3.1847)ds^(2)=(1+2Phi)dt^(2)-(1+2Phi)^(-1)dr^(2)-R^(2)dphi^(2):}\begin{equation*} d s^{2}=(1+2 \Phi) d t^{2}-(1+2 \Phi)^{-1} d r^{2}-R^{2} d \phi^{2} \tag{3.1847} \end{equation*}(3.1847)ds2=(1+2Φ)dt2(1+2Φ)1dr2R2dϕ2
The Lagrangian is given by
(3.1848) L = 1 = ( 1 + 2 Φ ) t ˙ 2 ( 1 + 2 Φ ) 1 r ˙ 2 R 2 ϕ ˙ 2 (3.1848) L = 1 = ( 1 + 2 Φ ) t ˙ 2 ( 1 + 2 Φ ) 1 r ˙ 2 R 2 ϕ ˙ 2 {:(3.1848)L=1=(1+2Phi)t^(˙)^(2)-(1+2Phi)^(-1)r^(˙)^(2)-R^(2)phi^(˙)^(2):}\begin{equation*} \mathcal{L}=1=(1+2 \Phi) \dot{t}^{2}-(1+2 \Phi)^{-1} \dot{r}^{2}-R^{2} \dot{\phi}^{2} \tag{3.1848} \end{equation*}(3.1848)L=1=(1+2Φ)t˙2(1+2Φ)1r˙2R2ϕ˙2
The Euler-Lagrange equations are then given by
(3.1849) r ¨ 2 G M r 2 ( 1 + 2 Φ ) 1 r ˙ 2 G M r 2 ( 1 + 2 Φ ) t ˙ 2 + r ( 1 + 2 Φ ) ϕ ˙ 2 = 0 (3.1850) d d τ ( t ˙ ) = 0 (3.1851) d d τ ( ϕ ˙ ) = 0 (3.1849) r ¨ 2 G M r 2 ( 1 + 2 Φ ) 1 r ˙ 2 G M r 2 ( 1 + 2 Φ ) t ˙ 2 + r ( 1 + 2 Φ ) ϕ ˙ 2 = 0 (3.1850) d d τ ( t ˙ ) = 0 (3.1851) d d τ ( ϕ ˙ ) = 0 {:[(3.1849)r^(¨)-(2GM)/(r^(2))(1+2Phi)^(-1)r^(˙)^(2)-(GM)/(r^(2))(1+2Phi)t^(˙)^(2)+r(1+2Phi)phi^(˙)^(2)=0],[(3.1850)(d)/(d tau)(t^(˙))=0],[(3.1851)(d)/(d tau)(phi^(˙))=0]:}\begin{align*} \ddot{r}-\frac{2 G M}{r^{2}}(1+2 \Phi)^{-1} \dot{r}^{2}-\frac{G M}{r^{2}}(1+2 \Phi) \dot{t}^{2}+r(1+2 \Phi) \dot{\phi}^{2} & =0 \tag{3.1849}\\ \frac{d}{d \tau}(\dot{t}) & =0 \tag{3.1850}\\ \frac{d}{d \tau}(\dot{\phi}) & =0 \tag{3.1851} \end{align*}(3.1849)r¨2GMr2(1+2Φ)1r˙2GMr2(1+2Φ)t˙2+r(1+2Φ)ϕ˙2=0(3.1850)ddτ(t˙)=0(3.1851)ddτ(ϕ˙)=0
The last two equations can be integrated to give
(3.1852) t ˙ = α , (3.1853) ϕ ˙ = β , (3.1852) t ˙ = α , (3.1853) ϕ ˙ = β , {:[(3.1852)t^(˙)=alpha","],[(3.1853)phi^(˙)=beta","]:}\begin{align*} \dot{t} & =\alpha, \tag{3.1852}\\ \dot{\phi} & =\beta, \tag{3.1853} \end{align*}(3.1852)t˙=α,(3.1853)ϕ˙=β,
where α α alpha\alphaα and β β beta\betaβ are constants. From the first equation, we can determine a relation between t ˙ t ˙ t^(˙)\dot{t}t˙ and ϕ ˙ ϕ ˙ phi^(˙)\dot{\phi}ϕ˙, using r ˙ = 0 r ˙ = 0 r^(˙)=0\dot{r}=0r˙=0 and r ¨ = 0 r ¨ = 0 r^(¨)=0\ddot{r}=0r¨=0. Inserting the expressions for t ˙ t ˙ t^(˙)\dot{t}t˙ and ϕ ˙ ϕ ˙ phi^(˙)\dot{\phi}ϕ˙, we get
(3.1854) R β 2 G M R 2 α 2 = 0 (3.1854) R β 2 G M R 2 α 2 = 0 {:(3.1854)Rbeta^(2)-(GM)/(R^(2))alpha^(2)=0:}\begin{equation*} R \beta^{2}-\frac{G M}{R^{2}} \alpha^{2}=0 \tag{3.1854} \end{equation*}(3.1854)Rβ2GMR2α2=0
Furthermore, the Lagrangian can be reduced, using r ˙ = 0 r ˙ = 0 r^(˙)=0\dot{r}=0r˙=0, to
(3.1855) L = 1 = ( 1 + 2 Φ ) α 2 R 2 β 2 (3.1855) L = 1 = ( 1 + 2 Φ ) α 2 R 2 β 2 {:(3.1855)L=1=(1+2Phi)alpha^(2)-R^(2)beta^(2):}\begin{equation*} \mathcal{L}=1=(1+2 \Phi) \alpha^{2}-R^{2} \beta^{2} \tag{3.1855} \end{equation*}(3.1855)L=1=(1+2Φ)α2R2β2
which can be solved for α = 1 1 + 3 2 Φ α = 1 1 + 3 2 Φ alpha=(1)/(sqrt(1+(3)/(2)*Phi))\alpha=\frac{1}{\sqrt{1+\frac{3}{2} \cdot \Phi}}α=11+32Φ and β = Φ R ( 2 + 3 Φ ) β = Φ R ( 2 + 3 Φ ) beta=sqrt((Phi)/(R(2+3Phi)))\beta=\sqrt{\frac{\Phi}{R(2+3 \Phi)}}β=ΦR(2+3Φ), and thus, from d ϕ d τ = β d ϕ d τ = β (d phi)/(d tau)=beta\frac{d \phi}{d \tau}=\betadϕdτ=β, we get
(3.1856) 2 π = β Δ τ Δ τ = 2 π R ( 2 + 3 Φ ) Φ (3.1856) 2 π = β Δ τ Δ τ = 2 π R ( 2 + 3 Φ ) Φ {:(3.1856)2pi=beta Delta tauquad=>quad Delta tau=2pisqrt((R(2+3Phi))/(Phi)):}\begin{equation*} 2 \pi=\beta \Delta \tau \quad \Rightarrow \quad \Delta \tau=2 \pi \sqrt{\frac{R(2+3 \Phi)}{\Phi}} \tag{3.1856} \end{equation*}(3.1856)2π=βΔτΔτ=2πR(2+3Φ)Φ
We can also determine t t ttt in terms of τ τ tau\tauτ as
(3.1857) Δ t = α Δ τ = Δ τ 1 + 3 2 Φ (3.1857) Δ t = α Δ τ = Δ τ 1 + 3 2 Φ {:(3.1857)Delta t=alpha Delta tau=(Delta tau)/(sqrt(1+(3)/(2)Phi)):}\begin{equation*} \Delta t=\alpha \Delta \tau=\frac{\Delta \tau}{\sqrt{1+\frac{3}{2} \Phi}} \tag{3.1857} \end{equation*}(3.1857)Δt=αΔτ=Δτ1+32Φ
Figure 3.17 Setup of the spherical body of radius R 0 R 0 R_(0)R_{0}R0 and impact parameter b b bbb and deflection angle α α alpha\alphaα of the neutrino, respectively.

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See Figure 3.17 for the setup of the problem. The deflection angle due to gravitational lensing in the weak-field limit is given by
(3.1858) α = | 2 ( ϕ ) d t | (3.1858) α = 2 ( ϕ ) d t {:(3.1858)alpha=|2int(grad phi)_(_|_)dt|:}\begin{equation*} \alpha=\left|2 \int(\nabla \phi)_{\perp} d t\right| \tag{3.1858} \end{equation*}(3.1858)α=|2(ϕ)dt|
where ϕ ϕ phi\phiϕ is the Newtonian gravitational potential and ( ϕ ) ( ϕ ) (grad phi)_(_|_)(\nabla \phi)_{\perp}(ϕ) is the component of its gradient which is perpendicular to the zeroth order direction of motion. We pick a coordinate system such that the zeroth order worldline of the neutrinos is given by x ( t ) = t e 1 + b e 2 x ( t ) = t e 1 + b e 2 x(t)=te_(1)+be_(2)x(t)=t e_{1}+b e_{2}x(t)=te1+be2. It follows that
(3.1859) α = 2 | e 2 ϕ d t | = 4 | 0 e 2 ϕ d t | (3.1859) α = 2 e 2 ϕ d t = 4 0 e 2 ϕ d t {:(3.1859)alpha=2|int_(-oo)^(oo)e_(2)*grad phi dt|=4|int_(0)^(oo)e_(2)*grad phi dt|:}\begin{equation*} \alpha=2\left|\int_{-\infty}^{\infty} e_{2} \cdot \nabla \phi d t\right|=4\left|\int_{0}^{\infty} e_{2} \cdot \nabla \phi d t\right| \tag{3.1859} \end{equation*}(3.1859)α=2|e2ϕdt|=4|0e2ϕdt|
The gradient ϕ ϕ grad phi\nabla \phiϕ is given by
(3.1860) ϕ = e r r ϕ = G M ( r ) r 2 e r (3.1860) ϕ = e r r ϕ = G M ( r ) r 2 e r {:(3.1860)grad phi=e_(r)del_(r)phi=(GM(r))/(r^(2))e_(r):}\begin{equation*} \nabla \phi=e_{r} \partial_{r} \phi=\frac{G M(r)}{r^{2}} e_{r} \tag{3.1860} \end{equation*}(3.1860)ϕ=errϕ=GM(r)r2er
where M ( r ) M ( r ) M(r)M(r)M(r) is the mass inside the radius r r rrr. This mass is given by
(3.1861) M ( r ) = { M 0 r 3 R 0 3 , r < r 0 M 0 , r r 0 (3.1861) M ( r ) = M 0 r 3 R 0 3 , r < r 0 M 0 , r r 0 {:(3.1861)M(r)={[M_(0)(r^(3))/(R_(0)^(3))","quad r < r_(0)],[M_(0)","quad r >= r_(0)]:}:}M(r)=\left\{\begin{array}{cc} M_{0} \frac{r^{3}}{R_{0}^{3}}, \quad r<r_{0} \tag{3.1861}\\ M_{0}, \quad r \geq r_{0} \end{array}\right.(3.1861)M(r)={M0r3R03,r<r0M0,rr0
Noting that r 2 = t 2 + b 2 r 2 = t 2 + b 2 r^(2)=t^(2)+b^(2)r^{2}=t^{2}+b^{2}r2=t2+b2, it follows that
0 e 2 ϕ d t = 0 G b M ( r ) r 3 d t = G M 0 0 R 0 2 b 2 b R 0 3 d t + G M 0 R 0 2 b 2 b r 3 d t = M 0 b G [ R 0 2 b 2 R 0 3 + 1 b 2 1 b 2 R 0 2 b 2 R 0 ] (3.1862) = M 0 b G [ R 0 2 b 2 R 0 ( 1 R 0 2 1 b 2 ) + 1 b 2 ] 0 e 2 ϕ d t = 0 G b M ( r ) r 3 d t = G M 0 0 R 0 2 b 2 b R 0 3 d t + G M 0 R 0 2 b 2 b r 3 d t = M 0 b G R 0 2 b 2 R 0 3 + 1 b 2 1 b 2 R 0 2 b 2 R 0 (3.1862) = M 0 b G R 0 2 b 2 R 0 1 R 0 2 1 b 2 + 1 b 2 {:[int_(0)^(oo)e_(2)*grad phi dt=int_(0)^(oo)(GbM(r))/(r^(3))dt=GM_(0)int_(0)^(sqrt(R_(0)^(2)-b^(2)))(b)/(R_(0)^(3))dt+GM_(0)int_(sqrt(R_(0)^(2)-b^(2)))^(oo)(b)/(r^(3))dt],[=M_(0)bG[(sqrt(R_(0)^(2)-b^(2)))/(R_(0)^(3))+(1)/(b^(2))-(1)/(b^(2))*(sqrt(R_(0)^(2)-b^(2)))/(R_(0))]],[(3.1862)=M_(0)bG[(sqrt(R_(0)^(2)-b^(2)))/(R_(0))((1)/(R_(0)^(2))-(1)/(b^(2)))+(1)/(b^(2))]]:}\begin{align*} \int_{0}^{\infty} e_{2} \cdot \nabla \phi d t & =\int_{0}^{\infty} \frac{G b M(r)}{r^{3}} d t=G M_{0} \int_{0}^{\sqrt{R_{0}^{2}-b^{2}}} \frac{b}{R_{0}^{3}} d t+G M_{0} \int_{\sqrt{R_{0}^{2}-b^{2}}}^{\infty} \frac{b}{r^{3}} d t \\ & =M_{0} b G\left[\frac{\sqrt{R_{0}^{2}-b^{2}}}{R_{0}^{3}}+\frac{1}{b^{2}}-\frac{1}{b^{2}} \cdot \frac{\sqrt{R_{0}^{2}-b^{2}}}{R_{0}}\right] \\ & =M_{0} b G\left[\frac{\sqrt{R_{0}^{2}-b^{2}}}{R_{0}}\left(\frac{1}{R_{0}^{2}}-\frac{1}{b^{2}}\right)+\frac{1}{b^{2}}\right] \tag{3.1862} \end{align*}0e2ϕdt=0GbM(r)r3dt=GM00R02b2bR03dt+GM0R02b2br3dt=M0bG[R02b2R03+1b21b2R02b2R0](3.1862)=M0bG[R02b2R0(1R021b2)+1b2]
This leads us to conclude that the deflection angle α α alpha\alphaα is given by
(3.1863) α = 4 M 0 b G [ R 0 2 b 2 R 0 ( 1 R 0 2 1 b 2 ) + 1 b 2 ] (3.1863) α = 4 M 0 b G R 0 2 b 2 R 0 1 R 0 2 1 b 2 + 1 b 2 {:(3.1863)alpha=4M_(0)bG[(sqrt(R_(0)^(2)-b^(2)))/(R_(0))((1)/(R_(0)^(2))-(1)/(b^(2)))+(1)/(b^(2))]:}\begin{equation*} \alpha=4 M_{0} b G\left[\frac{\sqrt{R_{0}^{2}-b^{2}}}{R_{0}}\left(\frac{1}{R_{0}^{2}}-\frac{1}{b^{2}}\right)+\frac{1}{b^{2}}\right] \tag{3.1863} \end{equation*}(3.1863)α=4M0bG[R02b2R0(1R021b2)+1b2]
For b R 0 , α 4 M 0 G b b R 0 , α 4 M 0 G b b rarrR_(0),alpha rarr(4M_(0)G)/(b)b \rightarrow R_{0}, \alpha \rightarrow \frac{4 M_{0} G}{b}bR0,α4M0Gb, which is the expected result as it coincides with the result for the deflection outside of a spherical mass distribution with total mass M 0 M 0 M_(0)M_{0}M0.
2.120
The Newtonian gravitational potential Φ Φ Phi\PhiΦ satisfies Poisson's equation
(3.1864) 2 Φ = 4 π G ρ (3.1864) 2 Φ = 4 π G ρ {:(3.1864)grad^(2)Phi=4pi G rho:}\begin{equation*} \nabla^{2} \Phi=4 \pi G \rho \tag{3.1864} \end{equation*}(3.1864)2Φ=4πGρ
in the region r > r s r > r s r > r_(s)r>r_{s}r>rs. Due to the symmetry of the problem, we will assume that Φ Φ Phi\PhiΦ is spherically symmetric and introduce the function f ( r ) = r Φ ( r ) f ( r ) = r Φ ( r ) f(r)=r Phi(r)f(r)=r \Phi(r)f(r)=rΦ(r), with which the differential equation for the gravitational potential takes the form
(3.1865) f ( r ) = 4 π r G ρ ( r ) (3.1865) f ( r ) = 4 π r G ρ ( r ) {:(3.1865)f^('')(r)=4pi rG rho(r):}\begin{equation*} f^{\prime \prime}(r)=4 \pi r G \rho(r) \tag{3.1865} \end{equation*}(3.1865)f(r)=4πrGρ(r)
With the assumption of the NFW halo profile for ρ ( r ) ρ ( r ) rho(r)\rho(r)ρ(r), we find that
(3.1866) f ( r ) = 4 π G k r 2 f ( r ) = 4 π G k ln ( r R ) + C r (3.1866) f ( r ) = 4 π G k r 2 f ( r ) = 4 π G k ln r R + C r {:(3.1866)f^('')(r)=(4pi Gk)/(r^(2))quad Longrightarrowquad f(r)=-4pi Gk ln((r)/(R))+Cr:}\begin{equation*} f^{\prime \prime}(r)=\frac{4 \pi G k}{r^{2}} \quad \Longrightarrow \quad f(r)=-4 \pi G k \ln \left(\frac{r}{R}\right)+C r \tag{3.1866} \end{equation*}(3.1866)f(r)=4πGkr2f(r)=4πGkln(rR)+Cr
where C C CCC and R R RRR are integration constants. The resulting expression for the gravitational potential is therefore
(3.1867) Φ ( r ) = 4 π G k r ln ( r R ) + C (3.1867) Φ ( r ) = 4 π G k r ln r R + C {:(3.1867)Phi(r)=-(4pi Gk)/(r)ln((r)/(R))+C:}\begin{equation*} \Phi(r)=-\frac{4 \pi G k}{r} \ln \left(\frac{r}{R}\right)+C \tag{3.1867} \end{equation*}(3.1867)Φ(r)=4πGkrln(rR)+C
The gradient of the potential is now
(3.1868) Φ = e r Φ ( r ) = 4 π G k r 2 ln ( r e R ) = e r [ 4 π G k r 2 ln ( r r s ) + K r 2 ] (3.1868) Φ = e r Φ ( r ) = 4 π G k r 2 ln r e R = e r 4 π G k r 2 ln r r s + K r 2 {:(3.1868)grad Phi=e_(r)Phi^(')(r)=(4pi Gk)/(r^(2))ln((r)/(eR))=e_(r)[(4pi Gk)/(r^(2))ln((r)/(r_(s)))+(K)/(r^(2))]:}\begin{equation*} \nabla \Phi=\mathbf{e}_{r} \Phi^{\prime}(r)=\frac{4 \pi G k}{r^{2}} \ln \left(\frac{r}{e R}\right)=\mathbf{e}_{r}\left[\frac{4 \pi G k}{r^{2}} \ln \left(\frac{r}{r_{s}}\right)+\frac{K}{r^{2}}\right] \tag{3.1868} \end{equation*}(3.1868)Φ=erΦ(r)=4πGkr2ln(reR)=er[4πGkr2ln(rrs)+Kr2]
where we have introduced the new constant K = 4 π G k ln ( r s / e R ) K = 4 π G k ln r s / e R K=4pi Gk ln(r_(s)//eR)K=4 \pi G k \ln \left(r_{s} / e R\right)K=4πGkln(rs/eR). The constant K K KKK can be related to the mass M 0 M 0 M_(0)M_{0}M0 enclosed within r < r s r < r s r < r_(s)r<r_{s}r<rs according to the relation
(3.1869) Φ ( r s ) = G M 0 r s 2 = K r s 2 K = G M 0 (3.1869) Φ r s = G M 0 r s 2 = K r s 2 K = G M 0 {:(3.1869)Phi^(')(r_(s))=(GM_(0))/(r_(s)^(2))=(K)/(r_(s)^(2))quad Longrightarrowquad K=GM_(0):}\begin{equation*} \Phi^{\prime}\left(r_{s}\right)=\frac{G M_{0}}{r_{s}^{2}}=\frac{K}{r_{s}^{2}} \quad \Longrightarrow \quad K=G M_{0} \tag{3.1869} \end{equation*}(3.1869)Φ(rs)=GM0rs2=Krs2K=GM0
Letting the zeroth order worldline of the light signal be given by x ( t ) = t e 1 + r 0 e 2 x ( t ) = t e 1 + r 0 e 2 x(t)=te_(1)+r_(0)e_(2)\mathbf{x}(t)=t \mathbf{e}_{1}+r_{0} \mathbf{e}_{2}x(t)=te1+r0e2, the first-order lensing is given by
(3.1870) θ 2 e 2 Φ d t = 2 G M 0 r 0 r 0 2 + t 2 d t + 4 π G k ln [ ( t 2 + r 0 2 ) / r s 2 ] r 0 2 + t 2 3 d t (3.1870) θ 2 e 2 Φ d t = 2 G M 0 r 0 r 0 2 + t 2 d t + 4 π G k ln t 2 + r 0 2 / r s 2 r 0 2 + t 2 3 d t {:(3.1870)theta≃2inte_(2)*grad Phi dt=2int_(-oo)^(oo)(GM_(0)r_(0))/(sqrt(r_(0)^(2)+t^(2)))dt+4pi Gkint_(-oo)^(oo)(ln[(t^(2)+r_(0)^(2))//r_(s)^(2)])/(sqrt(r_(0)^(2)+t^(2))^(3))dt:}\begin{equation*} \theta \simeq 2 \int \mathbf{e}_{2} \cdot \nabla \Phi d t=2 \int_{-\infty}^{\infty} \frac{G M_{0} r_{0}}{\sqrt{r_{0}^{2}+t^{2}}} d t+4 \pi G k \int_{-\infty}^{\infty} \frac{\ln \left[\left(t^{2}+r_{0}^{2}\right) / r_{s}^{2}\right]}{{\sqrt{r_{0}^{2}+t^{2}}}^{3}} d t \tag{3.1870} \end{equation*}(3.1870)θ2e2Φdt=2GM0r0r02+t2dt+4πGkln[(t2+r02)/rs2]r02+t23dt
The first of these integrals is the same integral that we have encountered for the gravitational lensing of a point source. The second integral can be solved through a lengthy process involving substitutions and partial integrations to give the final result
(3.1871) θ 4 G M 0 r 0 + 16 π G k r 0 ln ( e r 0 2 r s ) 4 G M 0 r 0 + 16 π G k r 0 [ ln ( r 0 r s ) + 0.31 ] (3.1871) θ 4 G M 0 r 0 + 16 π G k r 0 ln e r 0 2 r s 4 G M 0 r 0 + 16 π G k r 0 ln r 0 r s + 0.31 {:(3.1871)theta≃(4GM_(0))/(r_(0))+(16 pi Gk)/(r_(0))ln((er_(0))/(2r_(s)))≃(4GM_(0))/(r_(0))+(16 pi Gk)/(r_(0))[ln((r_(0))/(r_(s)))+0.31]:}\begin{equation*} \theta \simeq \frac{4 G M_{0}}{r_{0}}+\frac{16 \pi G k}{r_{0}} \ln \left(\frac{e r_{0}}{2 r_{s}}\right) \simeq \frac{4 G M_{0}}{r_{0}}+\frac{16 \pi G k}{r_{0}}\left[\ln \left(\frac{r_{0}}{r_{s}}\right)+0.31\right] \tag{3.1871} \end{equation*}(3.1871)θ4GM0r0+16πGkr0ln(er02rs)4GM0r0+16πGkr0[ln(r0rs)+0.31]
where e = 2.71828 e = 2.71828 e=2.71828 dotse=2.71828 \ldotse=2.71828 is the base of the natural logarithm.

2.121

a) Consider two observers A A AAA and B B BBB located at spacetime points x A μ x A μ x_(A)^(mu)x_{A}^{\mu}xAμ and x B μ x B μ x_(B)^(mu)x_{B}^{\mu}xBμ, respectively. Suppose that A A AAA is sending out light waves at the rate n n nnn per coordinate time interval Δ Δ Delta\DeltaΔ. In the (local) rest frame of an observer, the time interval Δ Δ Delta\DeltaΔ is related to the proper time interval by the factor g 00 g 00 sqrt(g_(00))\sqrt{g_{00}}g00. It follows that the relation between the number of light waves received by B B BBB per unit proper time and the number of light waves emitted by A A AAA per unit proper time is
(3.1872) v B v A = g 00 ( x A ) g 00 ( x B ) , λ B λ A = g 00 ( x B ) g 00 ( x A ) (3.1872) v B v A = g 00 x A g 00 x B , λ B λ A = g 00 x B g 00 x A {:(3.1872)(v_(B))/(v_(A))=sqrt((g_(00)(x_(A)))/(g_(00)(x_(B))))","quad(lambda_(B))/(lambda_(A))=sqrt((g_(00)(x_(B)))/(g_(00)(x_(A)))):}\begin{equation*} \frac{v_{B}}{v_{A}}=\sqrt{\frac{g_{00}\left(x_{A}\right)}{g_{00}\left(x_{B}\right)}}, \quad \frac{\lambda_{B}}{\lambda_{A}}=\sqrt{\frac{g_{00}\left(x_{B}\right)}{g_{00}\left(x_{A}\right)}} \tag{3.1872} \end{equation*}(3.1872)vBvA=g00(xA)g00(xB),λBλA=g00(xB)g00(xA)
where v A ( λ A ) v A λ A v_(A)(lambda_(A))v_{A}\left(\lambda_{A}\right)vA(λA) is the frequency (wavelength) of the light emitted by A A AAA and v B ( λ B ) v B λ B v_(B)(lambda_(B))v_{B}\left(\lambda_{B}\right)vB(λB) is the frequency (wavelength) of the light received by B B BBB. Note that a frequency v v vvv and its corresponding wavelength λ λ lambda\lambdaλ are related as v λ = c v λ = c v lambda=cv \lambda=cvλ=c. The light is traveling along a null geodesic (i.e., a geodesic such that the tangent vector at each point is lightlike) from A A AAA to B B BBB, but we do not need the explicit solution of the geodesic equations.
b) In general, if the radial coordinate of B B BBB is r B r r B r r_(B)≫r_(**)r_{B} \gg r_{*}rBr in the Schwarzschild spacetime with g 00 = 1 r / r g 00 = 1 r / r g_(00)=1-r_(**)//rg_{00}=1-r_{*} / rg00=1r/r, then
(3.1873) v B v A 1 r r A , λ B λ A 1 1 r r A (3.1873) v B v A 1 r r A , λ B λ A 1 1 r r A {:(3.1873)(v_(B))/(v_(A))≃sqrt(1-(r_(**))/(r_(A)))","quad(lambda_(B))/(lambda_(A))≃(1)/(sqrt(1-(r_(**))/(r_(A)))):}\begin{equation*} \frac{v_{B}}{v_{A}} \simeq \sqrt{1-\frac{r_{*}}{r_{A}}}, \quad \frac{\lambda_{B}}{\lambda_{A}} \simeq \frac{1}{\sqrt{1-\frac{r_{*}}{r_{A}}}} \tag{3.1873} \end{equation*}(3.1873)vBvA1rrA,λBλA11rrA
Furthermore, if it holds for the radial coordinate of A A AAA that r A r r A r r_(A)≫r_(**)r_{A} \gg r_{*}rAr, then
(3.1874) v B v A 1 1 2 r r A , λ B λ A 1 + 1 2 r r A (3.1874) v B v A 1 1 2 r r A , λ B λ A 1 + 1 2 r r A {:(3.1874)(v_(B))/(v_(A))≃1-(1)/(2)(r_(**))/(r_(A))","quad(lambda_(B))/(lambda_(A))≃1+(1)/(2)(r_(**))/(r_(A)):}\begin{equation*} \frac{v_{B}}{v_{A}} \simeq 1-\frac{1}{2} \frac{r_{*}}{r_{A}}, \quad \frac{\lambda_{B}}{\lambda_{A}} \simeq 1+\frac{1}{2} \frac{r_{*}}{r_{A}} \tag{3.1874} \end{equation*}(3.1874)vBvA112rrA,λBλA1+12rrA
This means that the frequency observed by B B BBB is actually smaller than the frequency emitted by A A AAA and the wavelength observed by B B BBB is longer than the wavelength emitted by A A AAA, i.e., the light is redshifted. The redshift z z zzz is defined as z λ B / λ A 1 z λ B / λ A 1 z-=lambda_(B)//lambda_(A)-1z \equiv \lambda_{B} / \lambda_{A}-1zλB/λA1.
Finally, assume that light is emitted by A A AAA at r A r A r_(A)r_{A}rA (e.g., from the surface of a star) and the light is observed by B B BBB far away (on Earth), i.e., r B r r r B r r r_(B)-=r_(oo)≫r_(**)r_{B} \equiv r_{\infty} \gg r_{*}rBrr, then the formula for the gravitational redshift is given by
(3.1875) z λ λ A 1 1 1 r r A 1 = 1 1 2 G M c 2 r A 1 (3.1875) z λ λ A 1 1 1 r r A 1 = 1 1 2 G M c 2 r A 1 {:(3.1875)z_(oo)-=(lambda_(oo))/(lambda_(A))-1≃(1)/(sqrt(1-(r_(**))/(r_(A))))-1=(1)/(sqrt(1-(2GM)/(c^(2)r_(A))))-1:}\begin{equation*} z_{\infty} \equiv \frac{\lambda_{\infty}}{\lambda_{A}}-1 \simeq \frac{1}{\sqrt{1-\frac{r_{*}}{r_{A}}}}-1=\frac{1}{\sqrt{1-\frac{2 G M}{c^{2} r_{A}}}}-1 \tag{3.1875} \end{equation*}(3.1875)zλλA111rrA1=112GMc2rA1
Note that for light emitted at r r r r r rarrr_(**)r \rightarrow r_{*}rr, the redshift grows to infinity, whereas for light emitted from r A r r A r r_(A)≫r_(**)r_{A} \gg r_{*}rAr, there is no redshift. In the Newtonian limit, when r A r A r_(A)r_{A}rA is sufficiently large compared to r r r_(**)r_{*}r, i.e., r A r r A r r_(A)≫r_(**)r_{A} \gg r_{*}rAr, the gravitational redshift can be approximated as
(3.1876) z 1 2 r r A = G M c 2 r A (3.1876) z 1 2 r r A = G M c 2 r A {:(3.1876)z_(oo)≃(1)/(2)(r_(**))/(r_(A))=(GM)/(c^(2)r_(A)):}\begin{equation*} z_{\infty} \simeq \frac{1}{2} \frac{r_{*}}{r_{A}}=\frac{G M}{c^{2} r_{A}} \tag{3.1876} \end{equation*}(3.1876)z12rrA=GMc2rA
White dwarfs like Sirius B and 40 Eridani B do show gravitational redshifts in the range between 10 4 10 4 10^(-4)10^{-4}104 and 10 5 10 5 10^(-5)10^{-5}105, which are of the right order of magnitude. More reliable and quantitatively accurate measurements are possible only in terrestrial experiments. For example, in 1960, Pound and Rebka measured the change of frequency of a γ γ gamma\gammaγ-ray photon emitted by an excited iron nucleus as it fell from a height of 18 21 m 18 21 m 18-21m18-21 \mathrm{~m}1821 m. When the photon falls from a height h h hhh, the change in the Newtonian gravitational potential is g h g h ghg hgh, where g g ggg is the acceleration due to gravity on the Earth's surface. Since g 00 1 g 00 1 g_(00)-1g_{00}-1g001 is approximately given by the Newtonian gravitational potential, the frequency increases by a factor 1 + g h / c 2 1 + g h / c 2 1+gh//c^(2)1+g h / c^{2}1+gh/c2. The fraction g h / c 2 g h / c 2 gh//c^(2)g h / c^{2}gh/c2 is small, about 10 15 10 15 10^(-15)10^{-15}1015, but it can still be measured, confirming the gravitational redshift effect.

2.122

The surface ( c t ) 2 x 2 y 2 = K 2 ( c t ) 2 x 2 y 2 = K 2 (ct)^(2)-x^(2)-y^(2)=-K^(2)(c t)^{2}-x^{2}-y^{2}=-K^{2}(ct)2x2y2=K2, where K > 0 K > 0 K > 0K>0K>0, can be written as ( c t ) 2 r 2 = ( c t ) 2 r 2 = (ct)^(2)-r^(2)=(c t)^{2}-r^{2}=(ct)2r2= K 2 K 2 -K^(2)-K^{2}K2 if we introduce polar coordinates such that ( t , x , y ) = ( t , r cos ϕ , r sin ϕ ) ( t , x , y ) = ( t , r cos ϕ , r sin ϕ ) (t,x,y)=(t,r cos phi,r sin phi)(t, x, y)=(t, r \cos \phi, r \sin \phi)(t,x,y)=(t,rcosϕ,rsinϕ). Thus, we have
(3.1877) r = c 2 t 2 + K 2 (3.1877) r = c 2 t 2 + K 2 {:(3.1877)r=sqrt(c^(2)t^(2)+K^(2)):}\begin{equation*} r=\sqrt{c^{2} t^{2}+K^{2}} \tag{3.1877} \end{equation*}(3.1877)r=c2t2+K2
Therefore, the position vector is ( t , x , y ) = ( t , c 2 t 2 + K 2 cos ϕ , c 2 t 2 + K 2 sin ϕ ) ( t , x , y ) = t , c 2 t 2 + K 2 cos ϕ , c 2 t 2 + K 2 sin ϕ (t,x,y)=(t,sqrt(c^(2)t^(2)+K^(2))cos phi,sqrt(c^(2)t^(2)+K^(2))sin phi)(t, x, y)=\left(t, \sqrt{c^{2} t^{2}+K^{2}} \cos \phi, \sqrt{c^{2} t^{2}+K^{2}} \sin \phi\right)(t,x,y)=(t,c2t2+K2cosϕ,c2t2+K2sinϕ). Differentiating the position vector, we obtain
(3.1878) u t ( t , x , y ) (3.1879) = ( 1 , c 2 t r cos ϕ , c 2 t r sin ϕ ) v ϕ ( t , x , y ) = ( 0 , r sin ϕ , r cos ϕ ) (3.1878) u t ( t , x , y ) (3.1879) = 1 , c 2 t r cos ϕ , c 2 t r sin ϕ v ϕ ( t , x , y ) = ( 0 , r sin ϕ , r cos ϕ ) {:[(3.1878)u-=(del)/(del t)(t","x","y)],[(3.1879)=(1,(c^(2)t)/(r)cos phi,(c^(2)t)/(r)sin phi)],[v-=(del)/(del phi)(t","x","y)=(0","-r sin phi","r cos phi)]:}\begin{align*} & \mathbf{u} \equiv \frac{\partial}{\partial t}(t, x, y) \tag{3.1878}\\ &=\left(1, \frac{c^{2} t}{r} \cos \phi, \frac{c^{2} t}{r} \sin \phi\right) \tag{3.1879}\\ & \mathbf{v} \equiv \frac{\partial}{\partial \phi}(t, x, y)=(0,-r \sin \phi, r \cos \phi) \end{align*}(3.1878)ut(t,x,y)(3.1879)=(1,c2trcosϕ,c2trsinϕ)vϕ(t,x,y)=(0,rsinϕ,rcosϕ)
Using u u u\mathbf{u}u and v v v\mathbf{v}v, we find the components of the metric as
(3.1880) g t t = u u = c 2 c 4 t 2 r 2 = c 2 K 2 c 2 t 2 + K 2 > 0 (3.1881) g ϕ ϕ = v v = r 2 = ( c 2 t 2 + K 2 ) < 0 (3.1882) g t ϕ = g ϕ t = u v = 0 (3.1880) g t t = u u = c 2 c 4 t 2 r 2 = c 2 K 2 c 2 t 2 + K 2 > 0 (3.1881) g ϕ ϕ = v v = r 2 = c 2 t 2 + K 2 < 0 (3.1882) g t ϕ = g ϕ t = u v = 0 {:[(3.1880)g_(tt)=u*u=c^(2)-(c^(4)t^(2))/(r^(2))=(c^(2)K^(2))/(c^(2)t^(2)+K^(2)) > 0],[(3.1881)g_(phi phi)=v*v=-r^(2)=-(c^(2)t^(2)+K^(2)) < 0],[(3.1882)g_(t phi)=g_(phi t)=u*v=0]:}\begin{align*} g_{t t} & =\mathbf{u} \cdot \mathbf{u}=c^{2}-\frac{c^{4} t^{2}}{r^{2}}=\frac{c^{2} K^{2}}{c^{2} t^{2}+K^{2}}>0 \tag{3.1880}\\ g_{\phi \phi} & =\mathbf{v} \cdot \mathbf{v}=-r^{2}=-\left(c^{2} t^{2}+K^{2}\right)<0 \tag{3.1881}\\ g_{t \phi} & =g_{\phi t}=\mathbf{u} \cdot \mathbf{v}=0 \tag{3.1882} \end{align*}(3.1880)gtt=uu=c2c4t2r2=c2K2c2t2+K2>0(3.1881)gϕϕ=vv=r2=(c2t2+K2)<0(3.1882)gtϕ=gϕt=uv=0
Since the metric components are independent of ϕ ϕ phi\phiϕ, the vector field ϕ ϕ del_(phi)\partial_{\phi}ϕ is a Killing vector field and with N = α t + β ϕ N = α t + β ϕ N=alphadel_(t)+betadel_(phi)N=\alpha \partial_{t}+\beta \partial_{\phi}N=αt+βϕ being the 4 -frequency of a light pulse sent from A A AAA at time coordinate t t ttt to B B BBB at time coordinate t t t^(')t^{\prime}t, this implies that
(3.1883) g ( ϕ , N ) = β g ϕ ϕ = k (3.1883) g ϕ , N = β g ϕ ϕ = k {:(3.1883)-g(del_(phi),N)=-betag_(phi phi)=k:}\begin{equation*} -g\left(\partial_{\phi}, N\right)=-\beta g_{\phi \phi}=k \tag{3.1883} \end{equation*}(3.1883)g(ϕ,N)=βgϕϕ=k
is a constant. Furthermore, as the 4 -frequency is a null vector, we find
(3.1884) g ( N , N ) = g t t α 2 + g ϕ ϕ β 2 = 0 α = β 2 g ϕ ϕ g t t = k g t t g ϕ ϕ (3.1884) g ( N , N ) = g t t α 2 + g ϕ ϕ β 2 = 0 α = β 2 g ϕ ϕ g t t = k g t t g ϕ ϕ {:(3.1884)g(N","N)=g_(tt)alpha^(2)+g_(phi phi)beta^(2)=0quad Longrightarrowquad alpha=sqrt((-beta^(2)g_(phi phi))/(g_(tt)))=(k)/(sqrt(-g_(tt)g_(phi phi))):}\begin{equation*} g(N, N)=g_{t t} \alpha^{2}+g_{\phi \phi} \beta^{2}=0 \quad \Longrightarrow \quad \alpha=\sqrt{\frac{-\beta^{2} g_{\phi \phi}}{g_{t t}}}=\frac{k}{\sqrt{-g_{t t} g_{\phi \phi}}} \tag{3.1884} \end{equation*}(3.1884)g(N,N)=gttα2+gϕϕβ2=0α=β2gϕϕgtt=kgttgϕϕ
For any comoving observer with constant ϕ ϕ phi\phiϕ, the 4 -velocity of the observer is given by V = γ t V = γ t V=gammadel_(t)V=\gamma \partial_{t}V=γt. Normalizing this to one we obtain
(3.1885) g ( V , V ) = γ 2 g t t = 1 γ = 1 g t t (3.1885) g ( V , V ) = γ 2 g t t = 1 γ = 1 g t t {:(3.1885)g(V","V)=gamma^(2)g_(tt)=1quad Longrightarrowquad gamma=(1)/(sqrt(g_(tt))):}\begin{equation*} g(V, V)=\gamma^{2} g_{t t}=1 \quad \Longrightarrow \quad \gamma=\frac{1}{\sqrt{g_{t t}}} \tag{3.1885} \end{equation*}(3.1885)g(V,V)=γ2gtt=1γ=1gtt
The frequency ω ω omega\omegaω measured by such an observer is given by the inner product of the 4 -frequency with the 4 -velocity, i.e.,
(3.1886) ω = g ( N , V ) = α γ g t t = k g t t g ϕ ϕ 1 g t t g t t = k g ϕ ϕ (3.1886) ω = g ( N , V ) = α γ g t t = k g t t g ϕ ϕ 1 g t t g t t = k g ϕ ϕ {:(3.1886)omega=g(N","V)=alpha gammag_(tt)=(k)/(sqrt(-g_(tt)g_(phi phi)))(1)/(sqrt(g_(tt)))g_(tt)=(k)/(sqrt(-g_(phi phi))):}\begin{equation*} \omega=g(N, V)=\alpha \gamma g_{t t}=\frac{k}{\sqrt{-g_{t t} g_{\phi \phi}}} \frac{1}{\sqrt{g_{t t}}} g_{t t}=\frac{k}{\sqrt{-g_{\phi \phi}}} \tag{3.1886} \end{equation*}(3.1886)ω=g(N,V)=αγgtt=kgttgϕϕ1gttgtt=kgϕϕ
It follows that the frequencies of the light sent from A A AAA at time t t ttt and received by B B BBB at time t t t^(')t^{\prime}t are related as
(3.1887) ω B ω A = k g ϕ ϕ ( t ) k g ϕ ϕ ( t ) = g ϕ ϕ ( t ) g ϕ ϕ ( t ) = c 2 t 2 + K 2 c 2 t 2 + K 2 (3.1887) ω B ω A = k g ϕ ϕ t k g ϕ ϕ ( t ) = g ϕ ϕ t g ϕ ϕ ( t ) = c 2 t 2 + K 2 c 2 t 2 + K 2 {:(3.1887)(omega_(B))/(omega_(A))=((k)/(sqrt(-g_(phi phi)(t^(')))))/((k)/(sqrt(-g_(phi phi)(t))))=sqrt((g_(phi phi)(t^(')))/(g_(phi phi)(t)))=sqrt((c^(2)t^('2)+K^(2))/(c^(2)t^(2)+K^(2))):}\begin{equation*} \frac{\omega_{B}}{\omega_{A}}=\frac{\frac{k}{\sqrt{-g_{\phi \phi}\left(t^{\prime}\right)}}}{\frac{k}{\sqrt{-g_{\phi \phi}(t)}}}=\sqrt{\frac{g_{\phi \phi}\left(t^{\prime}\right)}{g_{\phi \phi}(t)}}=\sqrt{\frac{c^{2} t^{\prime 2}+K^{2}}{c^{2} t^{2}+K^{2}}} \tag{3.1887} \end{equation*}(3.1887)ωBωA=kgϕϕ(t)kgϕϕ(t)=gϕϕ(t)gϕϕ(t)=c2t2+K2c2t2+K2

2.123

Assume that the light signal has 4-frequency N N NNN. Since t t del_(t)\partial_{t}t is a Killing vector field of the Schwarzschild metric, we know that
(3.1888) g ( N , t ) = g 00 N t = E (3.1888) g N , t = g 00 N t = E {:(3.1888)g(N,del_(t))=g_(00)N^(t)=E:}\begin{equation*} g\left(N, \partial_{t}\right)=g_{00} N^{t}=E \tag{3.1888} \end{equation*}(3.1888)g(N,t)=g00Nt=E
is a constant. Furthermore, since the 4 -frequency is a null vector, it holds that
(3.1889) g ( N , N ) = g 00 ( N t ) 2 + g r r ( N r ) 2 = E 2 g 00 + g r r ( N r ) 2 = 0 N r = E (3.1889) g ( N , N ) = g 00 N t 2 + g r r N r 2 = E 2 g 00 + g r r N r 2 = 0 N r = E {:(3.1889)g(N","N)=g_(00)(N^(t))^(2)+g_(rr)(N^(r))^(2)=(E^(2))/(g_(00))+g_(rr)(N^(r))^(2)=0quad LongrightarrowquadN^(r)=-E:}\begin{equation*} g(N, N)=g_{00}\left(N^{t}\right)^{2}+g_{r r}\left(N^{r}\right)^{2}=\frac{E^{2}}{g_{00}}+g_{r r}\left(N^{r}\right)^{2}=0 \quad \Longrightarrow \quad N^{r}=-E \tag{3.1889} \end{equation*}(3.1889)g(N,N)=g00(Nt)2+grr(Nr)2=E2g00+grr(Nr)2=0Nr=E
where we have used that g r r = 1 / g 00 g r r = 1 / g 00 g_(rr)=-1//g_(00)g_{r r}=-1 / g_{00}grr=1/g00 and that the light is moving radially toward smaller r r rrr. When the light signal is emitted from r r r rarr oor \rightarrow \inftyr, we have g 00 1 g 00 1 g_(00)rarr1g_{00} \rightarrow 1g001 and thus E = ω 0 E = ω 0 E=omega_(0)E=\omega_{0}E=ω0, where ω 0 ω 0 omega_(0)\omega_{0}ω0 is the frequency observed by a stationary observer at infinity.
The 4-frequency is therefore generally given by
(3.1890) N = E g 00 t E r (3.1890) N = E g 00 t E r {:(3.1890)N=(E)/(g_(00))del_(t)-Edel_(r):}\begin{equation*} N=\frac{E}{g_{00}} \partial_{t}-E \partial_{r} \tag{3.1890} \end{equation*}(3.1890)N=Eg00tEr
For the 4-velocity V V VVV of the space ship, we know that
(3.1891) V = d t d τ t + d r d τ r = d t d τ ( t + v c r ) (3.1891) V = d t d τ t + d r d τ r = d t d τ t + v c r {:(3.1891)V=(dt)/(d tau)del_(t)+(dr)/(d tau)del_(r)=(dt)/(d tau)(del_(t)+v_(c)del_(r)):}\begin{equation*} V=\frac{d t}{d \tau} \partial_{t}+\frac{d r}{d \tau} \partial_{r}=\frac{d t}{d \tau}\left(\partial_{t}+v_{c} \partial_{r}\right) \tag{3.1891} \end{equation*}(3.1891)V=dtdτt+drdτr=dtdτ(t+vcr)
where v c = d r / d t = 0.1 v c = d r / d t = 0.1 v_(c)=dr//dt=-0.1v_{c}=d r / d t=-0.1vc=dr/dt=0.1 is the coordinate velocity. Normalizing the 4 -velocity to one, we find that
(3.1892) 1 = g ( V , V ) = ( d t d τ ) 2 [ g 00 + g r r v c 2 ] d t d τ = 1 g 00 + g r r v c 2 (3.1892) 1 = g ( V , V ) = d t d τ 2 g 00 + g r r v c 2 d t d τ = 1 g 00 + g r r v c 2 {:(3.1892)1=g(V","V)=((dt)/(d tau))^(2)[g_(00)+g_(rr)v_(c)^(2)]quad Longrightarrowquad(dt)/(d tau)=(1)/(sqrt(g_(00)+g_(rr)v_(c)^(2))):}\begin{equation*} 1=g(V, V)=\left(\frac{d t}{d \tau}\right)^{2}\left[g_{00}+g_{r r} v_{c}^{2}\right] \quad \Longrightarrow \quad \frac{d t}{d \tau}=\frac{1}{\sqrt{g_{00}+g_{r r} v_{c}^{2}}} \tag{3.1892} \end{equation*}(3.1892)1=g(V,V)=(dtdτ)2[g00+grrvc2]dtdτ=1g00+grrvc2
The frequency observed by the space ship is now given by
(3.1893) ω = g ( N , V ) = g 00 E g 00 d t d τ g r r E v c d t d τ = ω 0 1 g r r v c g 00 + g r r v c 2 (3.1893) ω = g ( N , V ) = g 00 E g 00 d t d τ g r r E v c d t d τ = ω 0 1 g r r v c g 00 + g r r v c 2 {:(3.1893)omega=g(N","V)=g_(00)(E)/(g_(00))(dt)/(d tau)-g_(rr)Ev_(c)(dt)/(d tau)=omega_(0)(1-g_(rr)v_(c))/(sqrt(g_(00)+g_(rr)v_(c)^(2))):}\begin{equation*} \omega=g(N, V)=g_{00} \frac{E}{g_{00}} \frac{d t}{d \tau}-g_{r r} E v_{c} \frac{d t}{d \tau}=\omega_{0} \frac{1-g_{r r} v_{c}}{\sqrt{g_{00}+g_{r r} v_{c}^{2}}} \tag{3.1893} \end{equation*}(3.1893)ω=g(N,V)=g00Eg00dtdτgrrEvcdtdτ=ω01grrvcg00+grrvc2
Using that g r r = 1 / g 00 g r r = 1 / g 00 g_(rr)=-1//g_(00)g_{r r}=-1 / g_{00}grr=1/g00 now leads to
(3.1894) ω = ω 0 1 + v c / g 00 g 00 v c 2 / g 00 = ω 0 g 00 g 00 + v c g 00 2 v c 2 = ω 0 g 00 g 00 + v c g 00 v c (3.1894) ω = ω 0 1 + v c / g 00 g 00 v c 2 / g 00 = ω 0 g 00 g 00 + v c g 00 2 v c 2 = ω 0 g 00 g 00 + v c g 00 v c {:(3.1894)omega=omega_(0)(1+v_(c)//g_(00))/(sqrt(g_(00)-v_(c)^(2)//g_(00)))=(omega_(0))/(sqrt(g_(00)))(g_(00)+v_(c))/(sqrt(g_(00)^(2)-v_(c)^(2)))=(omega_(0))/(sqrt(g_(00)))sqrt((g_(00)+v_(c))/(g_(00)-v_(c))):}\begin{equation*} \omega=\omega_{0} \frac{1+v_{c} / g_{00}}{\sqrt{g_{00}-v_{c}^{2} / g_{00}}}=\frac{\omega_{0}}{\sqrt{g_{00}}} \frac{g_{00}+v_{c}}{\sqrt{g_{00}^{2}-v_{c}^{2}}}=\frac{\omega_{0}}{\sqrt{g_{00}}} \sqrt{\frac{g_{00}+v_{c}}{g_{00}-v_{c}}} \tag{3.1894} \end{equation*}(3.1894)ω=ω01+vc/g00g00vc2/g00=ω0g00g00+vcg002vc2=ω0g00g00+vcg00vc
Using that λ 1 / ω λ 1 / ω lambda prop1//omega\lambda \propto 1 / \omegaλ1/ω and inserting g 00 = 1 2 G M / r g 00 = 1 2 G M / r g_(00)=1-2GM//rg_{00}=1-2 G M / rg00=12GM/r the yields
(3.1895) λ = λ 0 1 2 G M r 1 2 G M r v c 1 2 G M r + v c (3.1895) λ = λ 0 1 2 G M r 1 2 G M r v c 1 2 G M r + v c {:(3.1895)lambda=lambda_(0)sqrt(1-(2GM)/(r))sqrt((1-(2GM)/(r)-v_(c))/(1-(2GM)/(r)+v_(c))):}\begin{equation*} \lambda=\lambda_{0} \sqrt{1-\frac{2 G M}{r}} \sqrt{\frac{1-\frac{2 G M}{r}-v_{c}}{1-\frac{2 G M}{r}+v_{c}}} \tag{3.1895} \end{equation*}(3.1895)λ=λ012GMr12GMrvc12GMr+vc
Inserting the values provided in the problem, we find that
(3.1896) λ 4420 (3.1896) λ 4420 {:(3.1896)lambda≃4420"Å":}\begin{equation*} \lambda \simeq 4420 \AA \tag{3.1896} \end{equation*}(3.1896)λ4420
2.124
The gravitational redshift observed far away for light emitted from the surface of a star (when the mass of the star is M = 2 10 30 kg M = 2 10 30 kg M=2*10^(30)kgM=2 \cdot 10^{30} \mathrm{~kg}M=21030 kg and r star = 7 10 8 m r star  = 7 10 8 m r_("star ")=7*10^(8)mr_{\text {star }}=7 \cdot 10^{8} \mathrm{~m}rstar =7108 m ) is given by
(3.1897) z = λ λ star λ star G M c 2 r star 2 10 6 (3.1897) z = λ λ star  λ star  G M c 2 r star  2 10 6 {:(3.1897)z=(lambda_(oo)-lambda_("star "))/(lambda_("star "))≃(GM)/(c^(2)r_("star "))~~2*10^(-6):}\begin{equation*} z=\frac{\lambda_{\infty}-\lambda_{\text {star }}}{\lambda_{\text {star }}} \simeq \frac{G M}{c^{2} r_{\text {star }}} \approx 2 \cdot 10^{-6} \tag{3.1897} \end{equation*}(3.1897)z=λλstar λstar GMc2rstar 2106
See Problem 2.121 for a derivation of the formula for the gravitational redshift.
2.125
The Schwarzschild metric is given by
(3.1898) d s 2 = ( 1 2 G M c 2 r ) ( d x 0 ) 2 ( 1 2 G M c 2 r ) 1 d r 2 r 2 d Ω 2 (3.1898) d s 2 = 1 2 G M c 2 r d x 0 2 1 2 G M c 2 r 1 d r 2 r 2 d Ω 2 {:(3.1898)ds^(2)=(1-(2GM)/(c^(2)r))(dx^(0))^(2)-(1-(2GM)/(c^(2)r))^(-1)dr^(2)-r^(2)dOmega^(2):}\begin{equation*} d s^{2}=\left(1-\frac{2 G M}{c^{2} r}\right)\left(d x^{0}\right)^{2}-\left(1-\frac{2 G M}{c^{2} r}\right)^{-1} d r^{2}-r^{2} d \Omega^{2} \tag{3.1898} \end{equation*}(3.1898)ds2=(12GMc2r)(dx0)2(12GMc2r)1dr2r2dΩ2
From this metric, one obtains
(3.1899) g 00 ( r ) = 1 2 G M c 2 r (3.1899) g 00 ( r ) = 1 2 G M c 2 r {:(3.1899)g_(00)(r)=1-(2GM)/(c^(2)r):}\begin{equation*} g_{00}(r)=1-\frac{2 G M}{c^{2} r} \tag{3.1899} \end{equation*}(3.1899)g00(r)=12GMc2r
Now, one finds the ratio between the observed frequency v v v^(')v^{\prime}v and the emitted frequency v v vvv as (see the discussion in the solution to Problem 2.121)
(3.1900) v v = g 00 ( r ) g 00 ( r ) = 1 2 G M c 2 r 1 2 G M c 2 r (3.1900) v v = g 00 ( r ) g 00 r = 1 2 G M c 2 r 1 2 G M c 2 r {:(3.1900)(v^('))/(v)=sqrt((g_(00)(r))/(g_(00)(r^('))))=sqrt((1-(2GM)/(c^(2)r))/(1-(2GM)/(c^(2)r^(')))):}\begin{equation*} \frac{v^{\prime}}{v}=\sqrt{\frac{g_{00}(r)}{g_{00}\left(r^{\prime}\right)}}=\sqrt{\frac{1-\frac{2 G M}{c^{2} r}}{1-\frac{2 G M}{c^{2} r^{\prime}}}} \tag{3.1900} \end{equation*}(3.1900)vv=g00(r)g00(r)=12GMc2r12GMc2r
where r r rrr is the solar radius and r r r^(')r^{\prime}r is the average Sun-Earth distance.
Then, since r r r r r^(')≫rr^{\prime} \gg rrr and r 2 G M / c 2 r 2 G M / c 2 r≫2GM//c^(2)r \gg 2 G M / c^{2}r2GM/c2, one has
(3.1901) v v 1 2 G M c 2 r 1 G M c 2 r (3.1901) v v 1 2 G M c 2 r 1 G M c 2 r {:(3.1901)(v^('))/(v)≃sqrt(1-(2GM)/(c^(2)r))≃1-(GM)/(c^(2)r):}\begin{equation*} \frac{v^{\prime}}{v} \simeq \sqrt{1-\frac{2 G M}{c^{2} r}} \simeq 1-\frac{G M}{c^{2} r} \tag{3.1901} \end{equation*}(3.1901)vv12GMc2r1GMc2r
Finally, one obtains the magnitude and sign of the relative frequency shift Δ v / v Δ v / v Delta v//v\Delta v / vΔv/v of the spectral line as
(3.1902) Δ v v v v v = v v 1 G M c 2 r 1.9 10 6 (3.1902) Δ v v v v v = v v 1 G M c 2 r 1.9 10 6 {:(3.1902)(Delta v)/(v)-=(v^(')-v)/(v)=(v^('))/(v)-1≃-(GM)/(c^(2)r)~~-1.9*10^(-6):}\begin{equation*} \frac{\Delta v}{v} \equiv \frac{v^{\prime}-v}{v}=\frac{v^{\prime}}{v}-1 \simeq-\frac{G M}{c^{2} r} \approx-1.9 \cdot 10^{-6} \tag{3.1902} \end{equation*}(3.1902)Δvvvvv=vv1GMc2r1.9106
In addition, one can note that the redshift z z zzz is defined as and given by
(3.1903) z Δ λ λ λ λ λ = λ λ 1 = { λ v = c } = v v 1 G M c 2 r 1.9 10 6 (3.1903) z Δ λ λ λ λ λ = λ λ 1 = { λ v = c } = v v 1 G M c 2 r 1.9 10 6 {:(3.1903)z-=(Delta lambda)/(lambda)-=(lambda^(')-lambda)/(lambda)=(lambda^('))/(lambda)-1={lambda v=c}=(v)/(v^('))-1≃(GM)/(c^(2)r)~~1.9*10^(-6):}\begin{equation*} z \equiv \frac{\Delta \lambda}{\lambda} \equiv \frac{\lambda^{\prime}-\lambda}{\lambda}=\frac{\lambda^{\prime}}{\lambda}-1=\{\lambda v=c\}=\frac{v}{v^{\prime}}-1 \simeq \frac{G M}{c^{2} r} \approx 1.9 \cdot 10^{-6} \tag{3.1903} \end{equation*}(3.1903)zΔλλλλλ=λλ1={λv=c}=vv1GMc2r1.9106

2.126

When the spaceship is sending the light signal to the Earth, there are two effects: One is the redshift due to the Doppler effect, the other is a blueshift due to the gravitational pull from the Earth.
Let the frequency of the emitted light be ν ν nu\nuν. The redshift is then given by the z z zzz-factor, which is
(3.1904) z r = v v obs 1 v c (3.1904) z r = v v obs 1 v c {:(3.1904)z_(r)=(v)/(v_(obs))-1≃(v)/(c):}\begin{equation*} z_{r}=\frac{v}{v_{\mathrm{obs}}}-1 \simeq \frac{v}{c} \tag{3.1904} \end{equation*}(3.1904)zr=vvobs1vc
where v obs v obs  v_("obs ")v_{\text {obs }}vobs  is the observed frequency and v v vvv is the velocity of the spaceship. Using v = 100 m / s v = 100 m / s v=100m//sv=100 \mathrm{~m} / \mathrm{s}v=100 m/s and c = 3 10 8 m / s c = 3 10 8 m / s c=3*10^(8)m//sc=3 \cdot 10^{8} \mathrm{~m} / \mathrm{s}c=3108 m/s, we obtain for the redshift z r 3.3 10 7 z r 3.3 10 7 z_(r)~~3.3*10^(-7)z_{r} \approx 3.3 \cdot 10^{-7}zr3.3107.
For the blueshift due the the mass of the Earth, we use the Schwarzschild metric.
Thus, we have
(3.1905) z b = v v obs 1 = g 00 ( R ) g 00 ( R + h ) 1 , g 00 ( r ) = 1 2 G M c 2 r (3.1905) z b = v v obs 1 = g 00 ( R ) g 00 ( R + h ) 1 , g 00 ( r ) = 1 2 G M c 2 r {:(3.1905)z_(b)=(v)/(v_(obs))-1=sqrt((g_(00)(R))/(g_(00)(R+h)))-1","quadg_(00)(r)=1-(2GM)/(c^(2)r):}\begin{equation*} z_{b}=\frac{v}{v_{\mathrm{obs}}}-1=\sqrt{\frac{g_{00}(R)}{g_{00}(R+h)}}-1, \quad g_{00}(r)=1-\frac{2 G M}{c^{2} r} \tag{3.1905} \end{equation*}(3.1905)zb=vvobs1=g00(R)g00(R+h)1,g00(r)=12GMc2r
where R R RRR is the radius of the Earth and h h hhh is the altitude of the spaceship. Since h / R h / R h//Rh / Rh/R ( 0.16 1 ) ( 0.16 1 ) (~~0.16≪1)(\approx 0.16 \ll 1)(0.161) is small, we obtain
(3.1906) z b G M c 2 ( 1 R 1 R + h ) G M h c 2 R 2 = g h c 2 (3.1906) z b G M c 2 1 R 1 R + h G M h c 2 R 2 = g h c 2 {:(3.1906)z_(b)≃-(GM)/(c^(2))((1)/(R)-(1)/(R+h))≃-(GMh)/(c^(2)R^(2))=-(gh)/(c^(2)):}\begin{equation*} z_{b} \simeq-\frac{G M}{c^{2}}\left(\frac{1}{R}-\frac{1}{R+h}\right) \simeq-\frac{G M h}{c^{2} R^{2}}=-\frac{g h}{c^{2}} \tag{3.1906} \end{equation*}(3.1906)zbGMc2(1R1R+h)GMhc2R2=ghc2
where g g ggg is the acceleration at the Earth, i.e., g 9.8 m / s 2 g 9.8 m / s 2 g≃9.8m//s^(2)g \simeq 9.8 \mathrm{~m} / \mathrm{s}^{2}g9.8 m/s2. Since h = 10 6 m h = 10 6 m h=10^(6)mh=10^{6} \mathrm{~m}h=106 m, we obtain for the blueshift z b 1.1 10 10 z b 1.1 10 10 z_(b)~~-1.1*10^(-10)z_{b} \approx-1.1 \cdot 10^{-10}zb1.11010.
In conclusion, the redshift is much larger (i.e., 3000 times) than the blueshift. Thus, in this case, the Doppler effect is the most important physical effect.
2.127
The gravitational potential of the Sun at the Earth is given by
(3.1907) ϕ = G M r , (3.1907) ϕ = G M r , {:(3.1907)phi_(o.)=-(GM_(o.))/(r)",":}\begin{equation*} \phi_{\odot}=-\frac{G M_{\odot}}{r}, \tag{3.1907} \end{equation*}(3.1907)ϕ=GMr,
where r r rrr is the distance between the Sun and the Earth. Thus, the gravitational potential is
(3.1908) ϕ = 1.3 10 20 m 3 / s 2 1.5 10 11 m 9 10 8 m 2 / s 2 (3.1908) ϕ = 1.3 10 20 m 3 / s 2 1.5 10 11 m 9 10 8 m 2 / s 2 {:(3.1908)phi_(o.)=-(1.3*10^(20)m^(3)//s^(2))/(1.5*10^(11)(m))≃-9*10^(8)m^(2)//s^(2):}\begin{equation*} \phi_{\odot}=-\frac{1.3 \cdot 10^{20} \mathrm{~m}^{3} / \mathrm{s}^{2}}{1.5 \cdot 10^{11} \mathrm{~m}} \simeq-9 \cdot 10^{8} \mathrm{~m}^{2} / \mathrm{s}^{2} \tag{3.1908} \end{equation*}(3.1908)ϕ=1.31020 m3/s21.51011 m9108 m2/s2
which is an order of magnitude larger than the gravitational potential of the Earth at its surface. Thus, we can neglect the influence of the gravitational field of the Earth. The relation between the frequencies at ( v ) v (v_(o+))\left(v_{\oplus}\right)(v) and far away from ( v ) v (v_(oo))\left(v_{\infty}\right)(v) the Earth will be
(3.1909) v v 1 + 2 ϕ c 2 1 + ϕ c 2 (3.1909) v v 1 + 2 ϕ c 2 1 + ϕ c 2 {:(3.1909)(v_(oo))/(v_(o+))≃sqrt(1+(2phi_(o.))/(c^(2)))≃1+(phi_(o.))/(c^(2)):}\begin{equation*} \frac{v_{\infty}}{v_{\oplus}} \simeq \sqrt{1+\frac{2 \phi_{\odot}}{c^{2}}} \simeq 1+\frac{\phi_{\odot}}{c^{2}} \tag{3.1909} \end{equation*}(3.1909)vv1+2ϕc21+ϕc2
for small ϕ / c 2 ϕ / c 2 phi_(o.)//c^(2)\phi_{\odot} / c^{2}ϕ/c2. See the general discussion of this relation in the solution to Problem 2.125. The redshift parameter z z zzz is given by
(3.1910) z = λ λ λ = v v 1 ϕ c 2 9 10 8 m 2 / s 2 ( 3 10 8 m / s ) 2 = 10 8 (3.1910) z = λ λ λ = v v 1 ϕ c 2 9 10 8 m 2 / s 2 3 10 8 m / s 2 = 10 8 {:(3.1910)z=(lambda_(o+)-lambda_(oo))/(lambda_(oo))=(v_(oo))/(v_(o+))-1≃(phi_(o.))/(c^(2))≃-(9*10^(8)m^(2)//s^(2))/((3*10^(8)(m)//s)^(2))=-10^(-8):}\begin{equation*} z=\frac{\lambda_{\oplus}-\lambda_{\infty}}{\lambda_{\infty}}=\frac{v_{\infty}}{v_{\oplus}}-1 \simeq \frac{\phi_{\odot}}{c^{2}} \simeq-\frac{9 \cdot 10^{8} \mathrm{~m}^{2} / \mathrm{s}^{2}}{\left(3 \cdot 10^{8} \mathrm{~m} / \mathrm{s}\right)^{2}}=-10^{-8} \tag{3.1910} \end{equation*}(3.1910)z=λλλ=vv1ϕc29108 m2/s2(3108 m/s)2=108
(i.e., a blueshift, since z < 0 z < 0 z < 0z<0z<0 ).

2.128

We call the free-falling observer A A AAA and the observer at infinity B B BBB. The motion of the free-falling observer is governed by the radial differential equation
(3.1911) E r ˙ 2 2 = 1 2 ( 1 r r ) (3.1911) E r ˙ 2 2 = 1 2 1 r r {:(3.1911)E-(r^(˙)^(2))/(2)=(1)/(2)(1-(r_(**))/(r)):}\begin{equation*} E-\frac{\dot{r}^{2}}{2}=\frac{1}{2}\left(1-\frac{r_{*}}{r}\right) \tag{3.1911} \end{equation*}(3.1911)Er˙22=12(1rr)
where E E EEE is a constant of motion given by 2 E = g ( t , U ) 2 E = g t , U sqrt(2E)=g(del_(t),U)\sqrt{2 E}=g\left(\partial_{t}, U\right)2E=g(t,U), where U U UUU is the 4 -velocity of the observer A A AAA. Note that the constant of motion g ( φ , U ) = 0 g φ , U = 0 g(del_(varphi),U)=0g\left(\partial_{\varphi}, U\right)=0g(φ,U)=0, since the motion of A A AAA is assumed to be purely radial. In order to have a velocity just large enough to escape to infinity, we require that r ˙ 0 r ˙ 0 r^(˙)rarr0\dot{r} \rightarrow 0r˙0 as r r r rarr oor \rightarrow \inftyr, leading to E = 1 / 2 E = 1 / 2 E=1//2E=1 / 2E=1/2 and therefore
(3.1912) r ˙ = r r (3.1912) r ˙ = r r {:(3.1912)r^(˙)=sqrt((r_(**))/(r)):}\begin{equation*} \dot{r}=\sqrt{\frac{r_{*}}{r}} \tag{3.1912} \end{equation*}(3.1912)r˙=rr
From the expression for E E EEE in terms of g ( t , U ) g t , U g(del_(t),U)g\left(\partial_{t}, U\right)g(t,U), we also find
(3.1913) 1 = g ( t , U ) = g t t t ˙ t ˙ = r r r (3.1913) 1 = g t , U = g t t t ˙ t ˙ = r r r {:(3.1913)1=g(del_(t),U)=g_(tt)t^(˙)quad Longrightarrowquadt^(˙)=(r)/(r-r_(**)):}\begin{equation*} 1=g\left(\partial_{t}, U\right)=g_{t t} \dot{t} \quad \Longrightarrow \quad \dot{t}=\frac{r}{r-r_{*}} \tag{3.1913} \end{equation*}(3.1913)1=g(t,U)=gttt˙t˙=rrr
The 4-velocity of A A AAA is therefore
(3.1914) U = r r r t + r r r (3.1914) U = r r r t + r r r {:(3.1914)U=(r)/(r-r_(**))del_(t)+sqrt((r_(**))/(r))del_(r):}\begin{equation*} U=\frac{r}{r-r_{*}} \partial_{t}+\sqrt{\frac{r_{*}}{r}} \partial_{r} \tag{3.1914} \end{equation*}(3.1914)U=rrrt+rrr
We now let the 4 -frequency of the light be given by
(3.1915) N = α t + β r (3.1915) N = α t + β r {:(3.1915)N=alphadel_(t)+betadel_(r):}\begin{equation*} N=\alpha \partial_{t}+\beta \partial_{r} \tag{3.1915} \end{equation*}(3.1915)N=αt+βr
Since the 4-frequency is lightlike, we must have
(3.1916) g ( N , N ) = α 2 r r r β 2 r r r = 0 β = α r r r (3.1916) g ( N , N ) = α 2 r r r β 2 r r r = 0 β = α r r r {:(3.1916)g(N","N)=alpha^(2)(r-r_(**))/(r)-beta^(2)(r)/(r-r_(**))=0quad Longrightarrowquad beta=alpha(r-r_(**))/(r):}\begin{equation*} g(N, N)=\alpha^{2} \frac{r-r_{*}}{r}-\beta^{2} \frac{r}{r-r_{*}}=0 \quad \Longrightarrow \quad \beta=\alpha \frac{r-r_{*}}{r} \tag{3.1916} \end{equation*}(3.1916)g(N,N)=α2rrrβ2rrr=0β=αrrr
The frequency emitted by A A AAA at r = r A r = r A r=r_(A)r=r_{A}r=rA is given by
(3.1917) f 0 = g ( N , U ) = g t t α A t ˙ + g r r r ˙ β A = α A ( 1 r r A ) (3.1917) f 0 = g ( N , U ) = g t t α A t ˙ + g r r r ˙ β A = α A 1 r r A {:(3.1917)f_(0)=g(N","U)=g_(tt)alpha_(A)t^(˙)+g_(rr)r^(˙)beta_(A)=alpha_(A)(1-sqrt((r_(**))/(r_(A)))):}\begin{equation*} f_{0}=g(N, U)=g_{t t} \alpha_{A} \dot{t}+g_{r r} \dot{r} \beta_{A}=\alpha_{A}\left(1-\sqrt{\frac{r_{*}}{r_{A}}}\right) \tag{3.1917} \end{equation*}(3.1917)f0=g(N,U)=gttαAt˙+grrr˙βA=αA(1rrA)
leading to
(3.1918) α A = f 0 1 R / r A (3.1918) α A = f 0 1 R / r A {:(3.1918)alpha_(A)=(f_(0))/(1-sqrt(R//r_(A))):}\begin{equation*} \alpha_{A}=\frac{f_{0}}{1-\sqrt{R / r_{A}}} \tag{3.1918} \end{equation*}(3.1918)αA=f01R/rA
Furthermore, we know that N N NNN is tangent to and parallel along the lightlike geodesic describing the worldline of the light signal. Since t t del_(t)\partial_{t}t is a Killing vector field of the Schwarzschild spacetime, we find that
(3.1919) g ( t , N ) = r r r α = r A r r A α A = r A r r A f 0 1 r / r A (3.1919) g t , N = r r r α = r A r r A α A = r A r r A f 0 1 r / r A {:(3.1919)g(del_(t),N)=(r-r_(**))/(r)*alpha=(r_(A)r_(**))/(r_(A))*alpha_(A)=(r_(A)-r_(**))/(r_(A))(f_(0))/(1-sqrt(r_(**)//r_(A))):}\begin{equation*} g\left(\partial_{t}, N\right)=\frac{r-r_{*}}{r} \cdot \alpha=\frac{r_{A} r_{*}}{r_{A}} \cdot \alpha_{A}=\frac{r_{A}-r_{*}}{r_{A}} \frac{f_{0}}{1-\sqrt{r_{*} / r_{A}}} \tag{3.1919} \end{equation*}(3.1919)g(t,N)=rrrα=rArrAαA=rArrAf01r/rA
is a constant. For observer B B BBB at rest at infinity, the 4 -velocity is V = t V = t V=del_(t)V=\partial_{t}V=t and the frequency observed by B B BBB is therefore
(3.1920) f = g ( V , N ) = g ( t , N ) = f 0 r A r A r 1 r / r A (3.1920) f = g ( V , N ) = g t , N = f 0 r A r A r 1 r / r A {:(3.1920)f=g(V","N)=g(del_(t),N)=(f_(0))/(r_(A))(r_(A)-r_(**))/(1-sqrt(r_(**)//r_(A))):}\begin{equation*} f=g(V, N)=g\left(\partial_{t}, N\right)=\frac{f_{0}}{r_{A}} \frac{r_{A}-r_{*}}{1-\sqrt{r_{*} / r_{A}}} \tag{3.1920} \end{equation*}(3.1920)f=g(V,N)=g(t,N)=f0rArAr1r/rA

2.129

Light signals follow geodesics and their 4-frequency N N NNN is proportional to the tangent of the affinely parametrized geodesic. Since K = t K = t K=del_(t)K=\partial_{t}K=t is a Killing vector field, we know that
(3.1921) F = g ( t , N ) = g t t N t = x 2 N t (3.1921) F = g t , N = g t t N t = x 2 N t {:(3.1921)F=g(del_(t),N)=g_(tt)N^(t)=x^(2)N^(t):}\begin{equation*} F=g\left(\partial_{t}, N\right)=g_{t t} N^{t}=x^{2} N^{t} \tag{3.1921} \end{equation*}(3.1921)F=g(t,N)=gttNt=x2Nt
is constant, which means that N t = F / x 2 N t = F / x 2 N^(t)=F//x^(2)N^{t}=F / x^{2}Nt=F/x2. Since the light signals follow a lightlike worldline:
(3.1922) g ( N , N ) = x 2 ( N t ) 2 ( N x ) 2 = F 2 x 2 ( N x ) 2 = 0 N x = + F x (3.1922) g ( N , N ) = x 2 N t 2 N x 2 = F 2 x 2 N x 2 = 0 N x = + F x {:(3.1922)g(N","N)=x^(2)(N^(t))^(2)-(N^(x))^(2)=(F^(2))/(x^(2))-(N^(x))^(2)=0quad=>quadN^(x)=+(F)/(x):}\begin{equation*} g(N, N)=x^{2}\left(N^{t}\right)^{2}-\left(N^{x}\right)^{2}=\frac{F^{2}}{x^{2}}-\left(N^{x}\right)^{2}=0 \quad \Rightarrow \quad N^{x}=+\frac{F}{x} \tag{3.1922} \end{equation*}(3.1922)g(N,N)=x2(Nt)2(Nx)2=F2x2(Nx)2=0Nx=+Fx
The falling observer is also following a geodesic, so we also know that
(3.1923) Q = g ( t , γ ˙ 0 ) = g t t t ˙ = x 2 t ˙ (3.1923) Q = g t , γ ˙ 0 = g t t t ˙ = x 2 t ˙ {:(3.1923)Q=g(del_(t),gamma^(˙)_(0))=g_(tt)t^(˙)=x^(2)t^(˙):}\begin{equation*} Q=g\left(\partial_{t}, \dot{\gamma}_{0}\right)=g_{t t} \dot{t}=x^{2} \dot{t} \tag{3.1923} \end{equation*}(3.1923)Q=g(t,γ˙0)=gttt˙=x2t˙
is constant along its worldline γ 0 γ 0 gamma_(0)\gamma_{0}γ0. Furthermore, g ( γ ˙ 0 , γ ˙ 0 ) = Q 2 x 2 x ˙ 2 = 1 g γ ˙ 0 , γ ˙ 0 = Q 2 x 2 x ˙ 2 = 1 g(gamma^(˙)_(0),gamma^(˙)_(0))=(Q^(2))/(x^(2))-x^(˙)^(2)=1g\left(\dot{\gamma}_{0}, \dot{\gamma}_{0}\right)=\frac{Q^{2}}{x^{2}}-\dot{x}^{2}=1g(γ˙0,γ˙0)=Q2x2x˙2=1. Since the observer falls from x = x 0 x = x 0 x=x_(0)x=x_{0}x=x0, we find that Q = + x 0 Q = + x 0 Q=+x_(0)Q=+x_{0}Q=+x0. We can conclude that t ˙ = x 0 x 2 t ˙ = x 0 x 2 t^(˙)=(x_(0))/(x^(2))\dot{t}=\frac{x_{0}}{x^{2}}t˙=x0x2, x ˙ = x 0 r x 2 1 x ˙ = x 0 r x 2 1 x^(˙)=-sqrt((x_(0)r)/(x^(2))-1)\dot{x}=-\sqrt{\frac{x_{0} r}{x^{2}}-1}x˙=x0rx21 for the falling observer. The emitted frequency f e f e f_(e)f_{e}fe is given by
f e = g ( γ ˙ 0 , N ) | x = x e = x e 2 t ˙ N t x ˙ N x = x e 2 x 0 x e 2 F x e 2 + x 0 2 x e 2 1 F x e (3.1924) = F x e 2 [ x 0 + x 0 2 x e 2 ] f e = g γ ˙ 0 , N x = x e = x e 2 t ˙ N t x ˙ N x = x e 2 x 0 x e 2 F x e 2 + x 0 2 x e 2 1 F x e (3.1924) = F x e 2 x 0 + x 0 2 x e 2 {:[f_(e)=g(gamma^(˙)_(0),N)|_(x=x_(e))=x_(e)^(2)t^(˙)N^(t)-x^(˙)N^(x)=x_(e)^(2)(x_(0))/(x_(e)^(2))(F)/(x_(e)^(2))+sqrt((x_(0)^(2))/(x_(e)^(2))-1)(F)/(x_(e))],[(3.1924)=(F)/(x_(e)^(2))[x_(0)+sqrt(x_(0)^(2)-x_(e)^(2))]]:}\begin{align*} f_{e}=\left.g\left(\dot{\gamma}_{0}, N\right)\right|_{x=x_{e}}=x_{e}^{2} \dot{t} N^{t}-\dot{x} N^{x} & =x_{e}^{2} \frac{x_{0}}{x_{e}^{2}} \frac{F}{x_{e}^{2}}+\sqrt{\frac{x_{0}^{2}}{x_{e}^{2}}-1} \frac{F}{x_{e}} \\ & =\frac{F}{x_{e}^{2}}\left[x_{0}+\sqrt{x_{0}^{2}-x_{e}^{2}}\right] \tag{3.1924} \end{align*}fe=g(γ˙0,N)|x=xe=xe2t˙Ntx˙Nx=xe2x0xe2Fxe2+x02xe21Fxe(3.1924)=Fxe2[x0+x02xe2]
The observer at x = x 1 x = x 1 x=x_(1)x=x_{1}x=x1 with worldline γ 1 γ 1 gamma_(1)\gamma_{1}γ1 has x ˙ = 0 x ˙ = 0 x^(˙)=0\dot{x}=0x˙=0 and therefore g ( γ ˙ 1 , γ ˙ 1 ) = g γ ˙ 1 , γ ˙ 1 = g(gamma^(˙)_(1),gamma^(˙)_(1))=g\left(\dot{\gamma}_{1}, \dot{\gamma}_{1}\right)=g(γ˙1,γ˙1)= x 1 2 t ˙ 2 = 1 x 1 2 t ˙ 2 = 1 x_(1)^(2)t^(˙)^(2)=1x_{1}^{2} \dot{t}^{2}=1x12t˙2=1, which means that t ˙ = x 1 1 t ˙ = x 1 1 t^(˙)=x_(1)^(-1)\dot{t}=x_{1}^{-1}t˙=x11. The observed frequency is therefore
(3.1925) f obs = g ( γ ˙ 1 , N ) = x 1 2 1 x 1 N t = F x 1 (3.1925) f obs = g γ ˙ 1 , N = x 1 2 1 x 1 N t = F x 1 {:(3.1925)f_(obs)=g(gamma^(˙)_(1),N)=x_(1)^(2)(1)/(x_(1))N^(t)=(F)/(x_(1)):}\begin{equation*} f_{\mathrm{obs}}=g\left(\dot{\gamma}_{1}, N\right)=x_{1}^{2} \frac{1}{x_{1}} N^{t}=\frac{F}{x_{1}} \tag{3.1925} \end{equation*}(3.1925)fobs=g(γ˙1,N)=x121x1Nt=Fx1
and the ratio f e / f obs f e / f obs  f_(e)//f_("obs ")f_{e} / f_{\text {obs }}fe/fobs  is, thus, given by
(3.1926) f e f obs = x 1 x e 2 [ x 0 + x 0 2 x e 2 ] (3.1926) f e f obs = x 1 x e 2 x 0 + x 0 2 x e 2 {:(3.1926)(f_(e))/(f_(obs))=(x_(1))/(x_(e)^(2))[x_(0)+sqrt(x_(0)^(2)-x_(e)^(2))]:}\begin{equation*} \frac{f_{e}}{f_{\mathrm{obs}}}=\frac{x_{1}}{x_{e}^{2}}\left[x_{0}+\sqrt{x_{0}^{2}-x_{e}^{2}}\right] \tag{3.1926} \end{equation*}(3.1926)fefobs=x1xe2[x0+x02xe2]
The redshift z z zzz is therefore
(3.1927) z = f e f obs 1 = x 1 x e 2 [ x 0 + x 0 2 x e 2 ] 1 (3.1927) z = f e f obs  1 = x 1 x e 2 x 0 + x 0 2 x e 2 1 {:(3.1927)z=(f_(e))/(f_("obs "))-1=(x_(1))/(x_(e)^(2))[x_(0)+sqrt(x_(0)^(2)-x_(e)^(2))]-1:}\begin{equation*} z=\frac{f_{e}}{f_{\text {obs }}}-1=\frac{x_{1}}{x_{e}^{2}}\left[x_{0}+\sqrt{x_{0}^{2}-x_{e}^{2}}\right]-1 \tag{3.1927} \end{equation*}(3.1927)z=fefobs 1=x1xe2[x0+x02xe2]1
2.130
Consider the Robertson-Walker metric for d Ω = 0 d Ω = 0 d Omega=0d \Omega=0dΩ=0, i.e.,
(3.1928) d s 2 = c 2 d t 2 S ( t ) 2 1 k r 2 d r 2 (3.1928) d s 2 = c 2 d t 2 S ( t ) 2 1 k r 2 d r 2 {:(3.1928)ds^(2)=c^(2)dt^(2)-(S(t)^(2))/(1-kr^(2))dr^(2):}\begin{equation*} d s^{2}=c^{2} d t^{2}-\frac{S(t)^{2}}{1-k r^{2}} d r^{2} \tag{3.1928} \end{equation*}(3.1928)ds2=c2dt2S(t)21kr2dr2
Using d s 2 = 0 d s 2 = 0 ds^(2)=0d s^{2}=0ds2=0, which holds for the path of a light signal, we find that
(3.1929) 0 = c 2 d t 2 S ( t ) 2 1 k r 2 d r 2 c S ( t ) d t = 1 1 k r 2 d r (3.1929) 0 = c 2 d t 2 S ( t ) 2 1 k r 2 d r 2 c S ( t ) d t = 1 1 k r 2 d r {:(3.1929)0=c^(2)dt^(2)-(S(t)^(2))/(1-kr^(2))dr^(2)quad=>quad(c)/(S(t))dt=(1)/(sqrt(1-kr^(2)))dr:}\begin{equation*} 0=c^{2} d t^{2}-\frac{S(t)^{2}}{1-k r^{2}} d r^{2} \quad \Rightarrow \quad \frac{c}{S(t)} d t=\frac{1}{\sqrt{1-k r^{2}}} d r \tag{3.1929} \end{equation*}(3.1929)0=c2dt2S(t)21kr2dr2cS(t)dt=11kr2dr
where we assumed that d r / d t > 0 d r / d t > 0 dr//dt > 0d r / d t>0dr/dt>0, i.e., propagation forward in time. Since the observers are at rest with respect to r r rrr, we must have
(3.1930) t 0 t 1 d t S ( t ) = t 0 + ϵ t 1 + ϵ d t S ( t ) (3.1930) t 0 t 1 d t S ( t ) = t 0 + ϵ t 1 + ϵ d t S ( t ) {:(3.1930)int_(t_(0))^(t_(1))(dt)/(S(t))=int_(t_(0)+epsilon)^(t_(1)+epsilon^('))(dt)/(S(t)):}\begin{equation*} \int_{t_{0}}^{t_{1}} \frac{d t}{S(t)}=\int_{t_{0}+\epsilon}^{t_{1}+\epsilon^{\prime}} \frac{d t}{S(t)} \tag{3.1930} \end{equation*}(3.1930)t0t1dtS(t)=t0+ϵt1+ϵdtS(t)
from which, in the limit ϵ 0 ϵ 0 epsilon rarr0\epsilon \rightarrow 0ϵ0, we obtain
(3.1931) ϵ S ( t 0 ) ϵ S ( t 1 ) (3.1931) ϵ S t 0 ϵ S t 1 {:(3.1931)(epsilon)/(S(t_(0)))≃(epsilon^('))/(S(t_(1))):}\begin{equation*} \frac{\epsilon}{S\left(t_{0}\right)} \simeq \frac{\epsilon^{\prime}}{S\left(t_{1}\right)} \tag{3.1931} \end{equation*}(3.1931)ϵS(t0)ϵS(t1)
Thus, the cosmological redshift z z zzz is given by
(3.1932) 1 + z ϵ ϵ = S ( t 0 ) S ( t 1 ) z = S ( t 0 ) S ( t 1 ) 1 (3.1932) 1 + z ϵ ϵ = S t 0 S t 1 z = S t 0 S t 1 1 {:(3.1932)1+z-=(epsilon)/(epsilon^('))=(S(t_(0)))/(S(t_(1)))quad=>quad z=(S(t_(0)))/(S(t_(1)))-1:}\begin{equation*} 1+z \equiv \frac{\epsilon}{\epsilon^{\prime}}=\frac{S\left(t_{0}\right)}{S\left(t_{1}\right)} \quad \Rightarrow \quad z=\frac{S\left(t_{0}\right)}{S\left(t_{1}\right)}-1 \tag{3.1932} \end{equation*}(3.1932)1+zϵϵ=S(t0)S(t1)z=S(t0)S(t1)1
2.131
The Robertson-Walker metric for k = 1 k = 1 k=1k=1k=1 is given by
(3.1933) d s 2 = c 2 d t 2 S ( t ) 2 [ d χ 2 + sin 2 χ ( d θ 2 + sin 2 θ d ϕ 2 ) ] (3.1933) d s 2 = c 2 d t 2 S ( t ) 2 d χ 2 + sin 2 χ d θ 2 + sin 2 θ d ϕ 2 {:(3.1933)ds^(2)=c^(2)dt^(2)-S(t)^(2)[dchi^(2)+sin^(2)chi(dtheta^(2)+sin^(2)theta dphi^(2))]:}\begin{equation*} d s^{2}=c^{2} d t^{2}-S(t)^{2}\left[d \chi^{2}+\sin ^{2} \chi\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right)\right] \tag{3.1933} \end{equation*}(3.1933)ds2=c2dt2S(t)2[dχ2+sin2χ(dθ2+sin2θdϕ2)]
Construct the Lagrangian
(3.1934) L = c 2 t ˙ 2 S ( t ) 2 [ χ ˙ 2 + sin 2 χ ( θ ˙ 2 + sin 2 θ ϕ ˙ 2 ) ] (3.1934) L = c 2 t ˙ 2 S ( t ) 2 χ ˙ 2 + sin 2 χ θ ˙ 2 + sin 2 θ ϕ ˙ 2 {:(3.1934)L=c^(2)t^(˙)^(2)-S(t)^(2)[chi^(˙)^(2)+sin^(2)chi(theta^(˙)^(2)+sin^(2)thetaphi^(˙)^(2))]:}\begin{equation*} \mathscr{L}=c^{2} \dot{t}^{2}-S(t)^{2}\left[\dot{\chi}^{2}+\sin ^{2} \chi\left(\dot{\theta}^{2}+\sin ^{2} \theta \dot{\phi}^{2}\right)\right] \tag{3.1934} \end{equation*}(3.1934)L=c2t˙2S(t)2[χ˙2+sin2χ(θ˙2+sin2θϕ˙2)]
where the dot indicates differentiation with respect to the path parameter s s sss. Using Euler-Lagrange equations yields the differential equations for the geodesics, i.e.,
(3.1935) c 2 t ¨ + S ( t ) S ( t ) [ χ ˙ 2 + sin 2 χ ( θ ˙ 2 + sin 2 θ ϕ ˙ 2 ) ] = 0 (3.1936) d d s [ S ( t ) 2 χ ˙ ] 1 2 S ( t ) 2 sin 2 χ ( θ ˙ 2 + sin 2 θ ϕ ˙ 2 ) = 0 (3.1937) d d s [ S ( t ) 2 sin 2 χ θ ˙ ] 1 2 S ( t ) 2 sin 2 θ sin 2 χ ϕ ˙ 2 = 0 (3.1938) d d s [ S ( t ) 2 sin 2 χ sin 2 θ ϕ ˙ ] = 0 (3.1935) c 2 t ¨ + S ( t ) S ( t ) χ ˙ 2 + sin 2 χ θ ˙ 2 + sin 2 θ ϕ ˙ 2 = 0 (3.1936) d d s S ( t ) 2 χ ˙ 1 2 S ( t ) 2 sin 2 χ θ ˙ 2 + sin 2 θ ϕ ˙ 2 = 0 (3.1937) d d s S ( t ) 2 sin 2 χ θ ˙ 1 2 S ( t ) 2 sin 2 θ sin 2 χ ϕ ˙ 2 = 0 (3.1938) d d s S ( t ) 2 sin 2 χ sin 2 θ ϕ ˙ = 0 {:[(3.1935)c^(2)t^(¨)+S(t)S^(')(t)[chi^(˙)^(2)+sin^(2)chi(theta^(˙)^(2)+sin^(2)thetaphi^(˙)^(2))]=0],[(3.1936)(d)/(ds)[S(t)^(2)(chi^(˙))]-(1)/(2)S(t)^(2)sin 2chi(theta^(˙)^(2)+sin^(2)thetaphi^(˙)^(2))=0],[(3.1937)(d)/(ds)[S(t)^(2)sin^(2)chi(theta^(˙))]-(1)/(2)S(t)^(2)sin 2thetasin^(2)chiphi^(˙)^(2)=0],[(3.1938)(d)/(ds)[S(t)^(2)sin^(2)chisin^(2)theta(phi^(˙))]=0]:}\begin{align*} c^{2} \ddot{t}+S(t) S^{\prime}(t)\left[\dot{\chi}^{2}+\sin ^{2} \chi\left(\dot{\theta}^{2}+\sin ^{2} \theta \dot{\phi}^{2}\right)\right] & =0 \tag{3.1935}\\ \frac{d}{d s}\left[S(t)^{2} \dot{\chi}\right]-\frac{1}{2} S(t)^{2} \sin 2 \chi\left(\dot{\theta}^{2}+\sin ^{2} \theta \dot{\phi}^{2}\right) & =0 \tag{3.1936}\\ \frac{d}{d s}\left[S(t)^{2} \sin ^{2} \chi \dot{\theta}\right]-\frac{1}{2} S(t)^{2} \sin 2 \theta \sin ^{2} \chi \dot{\phi}^{2} & =0 \tag{3.1937}\\ \frac{d}{d s}\left[S(t)^{2} \sin ^{2} \chi \sin ^{2} \theta \dot{\phi}\right] & =0 \tag{3.1938} \end{align*}(3.1935)c2t¨+S(t)S(t)[χ˙2+sin2χ(θ˙2+sin2θϕ˙2)]=0(3.1936)dds[S(t)2χ˙]12S(t)2sin2χ(θ˙2+sin2θϕ˙2)=0(3.1937)dds[S(t)2sin2χθ˙]12S(t)2sin2θsin2χϕ˙2=0(3.1938)dds[S(t)2sin2χsin2θϕ˙]=0

2.132

a) The metric in the χ χ chi^(')\chi^{\prime}χ coordinates becomes
g a b = χ c χ a χ d χ b g c d ( δ a c ε a ξ c ) ( δ b d ε b ξ d ) ( η c d + ε h c d ) (3.1939) η a b + ε ( h a b a ξ b b ξ a ) , g a b = χ c χ a χ d χ b g c d δ a c ε a ξ c δ b d ε b ξ d η c d + ε h c d (3.1939) η a b + ε h a b a ξ b b ξ a , {:[g_(ab)^(')=(delchi^(c))/(delchi^('a))(delchi^(d))/(delchi^('b))g_(cd)≃(delta_(a)^(c)-epsidel_(a)xi^(c))(delta_(b)^(d)-epsidel_(b)xi^(d))(eta_(cd)+epsih_(cd))],[(3.1939)≃eta_(ab)+epsi(h_(ab)-del_(a)xi_(b)-del_(b)xi_(a))","]:}\begin{align*} g_{a b}^{\prime}=\frac{\partial \chi^{c}}{\partial \chi^{\prime a}} \frac{\partial \chi^{d}}{\partial \chi^{\prime b}} g_{c d} & \simeq\left(\delta_{a}^{c}-\varepsilon \partial_{a} \xi^{c}\right)\left(\delta_{b}^{d}-\varepsilon \partial_{b} \xi^{d}\right)\left(\eta_{c d}+\varepsilon h_{c d}\right) \\ & \simeq \eta_{a b}+\varepsilon\left(h_{a b}-\partial_{a} \xi_{b}-\partial_{b} \xi_{a}\right), \tag{3.1939} \end{align*}gab=χcχaχdχbgcd(δacεaξc)(δbdεbξd)(ηcd+εhcd)(3.1939)ηab+ε(habaξbbξa),
where \simeq denotes equality up to linear order in ε ε epsi\varepsilonε. This is equivalent to
(3.1940) g a b = η a b + ε h a b (3.1940) g a b = η a b + ε h a b {:(3.1940)g_(ab)^(')=eta_(ab)+epsih_(ab)^('):}\begin{equation*} g_{a b}^{\prime}=\eta_{a b}+\varepsilon h_{a b}^{\prime} \tag{3.1940} \end{equation*}(3.1940)gab=ηab+εhab
with
(3.1941) h a b = h a b a ξ b b ξ a (3.1941) h a b = h a b a ξ b b ξ a {:(3.1941)h_(ab)^(')=h_(ab)-del_(a)xi_(b)-del_(b)xi_(a):}\begin{equation*} h_{a b}^{\prime}=h_{a b}-\partial_{a} \xi_{b}-\partial_{b} \xi_{a} \tag{3.1941} \end{equation*}(3.1941)hab=habaξbbξa
Note that, since the term including the perturbations is linear in ε ε epsi\varepsilonε already, the indices of ξ a ξ a xi^(a)\xi^{a}ξa are can be raised and lowered with the Minkowski metric without affecting this statement and the lowering of the indices inside the partial derivatives is consistent to linear order in ε ε epsi\varepsilonε. Thus, we are still in the weak-field regime after the coordinate change with the new metric perturbation given above. Furthermore, we find that
(3.1942) g a b a b χ c = g a b a b ( χ c + ε ξ c ) = 0 (3.1942) g a b a b χ c = g a b a b χ c + ε ξ c = 0 {:(3.1942)g^(ab)grad_(a)grad_(b)chi^('c)=g^(ab)grad_(a)grad_(b)(chi^(c)+epsixi^(c))=0:}\begin{equation*} g^{a b} \nabla_{a} \nabla_{b} \chi^{\prime c}=g^{a b} \nabla_{a} \nabla_{b}\left(\chi^{c}+\varepsilon \xi^{c}\right)=0 \tag{3.1942} \end{equation*}(3.1942)gababχc=gabab(χc+εξc)=0
as long as both χ c χ c chi^(c)\chi^{c}χc and ξ c ξ c xi^(c)\xi^{c}ξc are harmonic functions. The coordinate change therefore also preserves the harmonic gauge condition.
b) In the harmonic gauge, we find that
(3.1943) g a b a b χ c = g a b a δ b c = g a b Γ a b c = 0 , (3.1943) g a b a b χ c = g a b a δ b c = g a b Γ a b c = 0 , {:(3.1943)g^(ab)grad_(a)grad_(b)chi^(c)=g^(ab)grad_(a)delta_(b)^(c)=g^(ab)Gamma_(ab)^(c)=0",":}\begin{equation*} g^{a b} \nabla_{a} \nabla_{b} \chi^{c}=g^{a b} \nabla_{a} \delta_{b}^{c}=g^{a b} \Gamma_{a b}^{c}=0, \tag{3.1943} \end{equation*}(3.1943)gababχc=gabaδbc=gabΓabc=0,
where we note that b χ c = b χ c = δ b c b χ c = b χ c = δ b c grad_(b)chi^(c)=del_(b)chi^(c)=delta_(b)^(c)\nabla_{b} \chi^{c}=\partial_{b} \chi^{c}=\delta_{b}^{c}bχc=bχc=δbc and c c ccc is just a counter for the coordinate functions, not a tensor index. In the weak-field limit, we also have
(3.1944) Γ a b c ε 2 η c d ( a h d b + b h a d d h a b ) = ε 2 ( a h b c + b h a c c h a b ) (3.1944) Γ a b c ε 2 η c d a h d b + b h a d d h a b = ε 2 a h b c + b h a c c h a b {:(3.1944)Gamma_(ab)^(c)≃(epsi)/(2)eta^(cd)(del_(a)h_(db)+del_(b)h_(ad)-del_(d)h_(ab))=(epsi)/(2)(del_(a)h_(b)^(c)+del_(b)h_(a)^(c)-del^(c)h_(ab)):}\begin{equation*} \Gamma_{a b}^{c} \simeq \frac{\varepsilon}{2} \eta^{c d}\left(\partial_{a} h_{d b}+\partial_{b} h_{a d}-\partial_{d} h_{a b}\right)=\frac{\varepsilon}{2}\left(\partial_{a} h_{b}^{c}+\partial_{b} h_{a}^{c}-\partial^{c} h_{a b}\right) \tag{3.1944} \end{equation*}(3.1944)Γabcε2ηcd(ahdb+bhaddhab)=ε2(ahbc+bhacchab)
Contracting this with g a b g a b g^(ab)g^{a b}gab now leads to
g a b Γ a b c ε 2 η a b ( a h b c + b h a c c h a b ) = ε ( a h a c 1 2 c h ) = ε a ( h a c 1 2 η a c h ) (3.1945) = ε a h ¯ a c = 0 g a b Γ a b c ε 2 η a b a h b c + b h a c c h a b = ε a h a c 1 2 c h = ε a h a c 1 2 η a c h (3.1945) = ε a h ¯ a c = 0 {:[g^(ab)Gamma_(ab)^(c)≃(epsi)/(2)eta^(ab)(del_(a)h_(b)^(c)+del_(b)h_(a)^(c)-del^(c)h_(ab))],[=epsi(del_(a)h^(ac)-(1)/(2)del^(c)h)=epsidel_(a)(h^(ac)-(1)/(2)eta^(ac)h)],[(3.1945)=epsidel_(a) bar(h)^(ac)=0]:}\begin{align*} g^{a b} \Gamma_{a b}^{c} & \simeq \frac{\varepsilon}{2} \eta^{a b}\left(\partial_{a} h_{b}^{c}+\partial_{b} h_{a}^{c}-\partial^{c} h_{a b}\right) \\ & =\varepsilon\left(\partial_{a} h^{a c}-\frac{1}{2} \partial^{c} h\right)=\varepsilon \partial_{a}\left(h^{a c}-\frac{1}{2} \eta^{a c} h\right) \\ & =\varepsilon \partial_{a} \bar{h}^{a c}=0 \tag{3.1945} \end{align*}gabΓabcε2ηab(ahbc+bhacchab)=ε(ahac12ch)=εa(hac12ηach)(3.1945)=εah¯ac=0
By lowering the index c c ccc follows that a h ¯ a c = 0 a h ¯ a c = 0 del^(a) bar(h)_(ac)=0\partial^{a} \bar{h}_{a c}=0ah¯ac=0.
c) From a) follows that
(3.1946) h ¯ a b = h ¯ a b a ξ b b ξ a + η a b c ξ c (3.1946) h ¯ a b = h ¯ a b a ξ b b ξ a + η a b c ξ c {:(3.1946) bar(h)_(ab)^(')= bar(h)_(ab)-del_(a)xi_(b)-del_(b)xi_(a)+eta_(ab)del_(c)xi^(c):}\begin{equation*} \bar{h}_{a b}^{\prime}=\bar{h}_{a b}-\partial_{a} \xi_{b}-\partial_{b} \xi_{a}+\eta_{a b} \partial_{c} \xi^{c} \tag{3.1946} \end{equation*}(3.1946)h¯ab=h¯abaξbbξa+ηabcξc
With ξ a = i C a exp ( i k χ ) ξ a = i C a exp ( i k χ ) xi^(a)=iC^(a)exp(ik*chi)\xi^{a}=i C^{a} \exp (i k \cdot \chi)ξa=iCaexp(ikχ), where k χ = k c χ c k χ = k c χ c k*chi=k_(c)chi^(c)k \cdot \chi=k_{c} \chi^{c}kχ=kcχc, the partial derivatives of ξ a ξ a xi_(a)\xi_{a}ξa are given by
(3.1947) b ξ a = η a c b ξ c = η a c i C c i k b exp ( i k χ ) = C a k b exp ( i k χ ) (3.1947) b ξ a = η a c b ξ c = η a c i C c i k b exp ( i k χ ) = C a k b exp ( i k χ ) {:(3.1947)del_(b)xi_(a)=eta_(ac)del_(b)xi^(c)=eta_(ac)iC^(c)ik_(b)exp(ik*chi)=-C_(a)k_(b)exp(ik*chi):}\begin{equation*} \partial_{b} \xi_{a}=\eta_{a c} \partial_{b} \xi^{c}=\eta_{a c} i C^{c} i k_{b} \exp (i k \cdot \chi)=-C_{a} k_{b} \exp (i k \cdot \chi) \tag{3.1947} \end{equation*}(3.1947)bξa=ηacbξc=ηaciCcikbexp(ikχ)=Cakbexp(ikχ)
and thus, we obtain
(3.1948) A a b = A a b + C a k b + k a C b η a b k c C c (3.1948) A a b = A a b + C a k b + k a C b η a b k c C c {:(3.1948)A_(ab)^(')=A_(ab)+C_(a)k_(b)+k_(a)C_(b)-eta_(ab)k_(c)C^(c):}\begin{equation*} A_{a b}^{\prime}=A_{a b}+C_{a} k_{b}+k_{a} C_{b}-\eta_{a b} k_{c} C^{c} \tag{3.1948} \end{equation*}(3.1948)Aab=Aab+Cakb+kaCbηabkcCc
d) From the harmonic gauge condition, we find that
(3.1949) a h ¯ a b = A a b i k a exp ( i k χ ) = 0 (3.1949) a h ¯ a b = A a b i k a exp ( i k χ ) = 0 {:(3.1949)del^(a) bar(h)_(ab)=A_(ab)ik^(a)exp(ik*chi)=0:}\begin{equation*} \partial^{a} \bar{h}_{a b}=A_{a b} i k^{a} \exp (i k \cdot \chi)=0 \tag{3.1949} \end{equation*}(3.1949)ah¯ab=Aabikaexp(ikχ)=0
i.e., with the given k 0 = k 3 = 1 k 0 = k 3 = 1 k_(0)=k_(3)=1k_{0}=k_{3}=1k0=k3=1 (or, equivalently, k 0 = k 3 = 1 k 0 = k 3 = 1 k^(0)=-k^(3)=1k^{0}=-k^{3}=1k0=k3=1 ),
(3.1950) A 0 b k 0 + A 3 b k 3 = A 0 b A 3 b = 0 A 0 b = A 3 b (3.1950) A 0 b k 0 + A 3 b k 3 = A 0 b A 3 b = 0 A 0 b = A 3 b {:(3.1950)A_(0b)k^(0)+A_(3b)k^(3)=A_(0b)-A_(3b)=0quad LongrightarrowquadA_(0b)=A_(3b):}\begin{equation*} A_{0 b} k^{0}+A_{3 b} k^{3}=A_{0 b}-A_{3 b}=0 \quad \Longrightarrow \quad A_{0 b}=A_{3 b} \tag{3.1950} \end{equation*}(3.1950)A0bk0+A3bk3=A0bA3b=0A0b=A3b
for all b b bbb. Due to the symmetry of A A AAA, this also means that A b 0 = A b 3 A b 0 = A b 3 A_(b0)=A_(b3)A_{b 0}=A_{b 3}Ab0=Ab3. This leads to A 00 = A 03 = A 30 = A 33 A 00 = A 03 = A 30 = A 33 A_(00)^(')=A_(03)^(')=A_(30)^(')=A_(33)^(')A_{00}^{\prime}=A_{03}^{\prime}=A_{30}^{\prime}=A_{33}^{\prime}A00=A03=A30=A33 and it is therefore sufficient to require that A 00 = 0 A 00 = 0 A_(00)^(')=0A_{00}^{\prime}=0A00=0. Furthermore, we also find that
(3.1953) A 00 = A 00 + 2 C 0 C 0 + C 3 = A 00 + C 0 + C 3 = 0 ( i 3 ) A 0 i = A 0 i + C 0 k i + k 0 C i = A 0 i + C i = 0 A a a = A a a 2 k C = A a a 2 ( C 0 C 3 ) = 0 (3.1953) A 00 = A 00 + 2 C 0 C 0 + C 3 = A 00 + C 0 + C 3 = 0 ( i 3 ) A 0 i = A 0 i + C 0 k i + k 0 C i = A 0 i + C i = 0 A a a = A a a 2 k C = A a a 2 C 0 C 3 = 0 {:(3.1953){:[,A_(00)^(')=A_(00)+2C_(0)-C_(0)+C_(3)=A_(00)+C_(0)+C_(3)=0],[(i!=3)quad,A_(0i)^(')=A_(0i)+C_(0)k_(i)+k_(0)C_(i)=A_(0i)+C_(i)=0],[,A_(a)^('a)=A_(a)^(a)-2k*C=A_(a)^(a)-2(C_(0)-C_(3))=0]:}:}\begin{array}{ll} & A_{00}^{\prime}=A_{00}+2 C_{0}-C_{0}+C_{3}=A_{00}+C_{0}+C_{3}=0 \\ (i \neq 3) \quad & A_{0 i}^{\prime}=A_{0 i}+C_{0} k_{i}+k_{0} C_{i}=A_{0 i}+C_{i}=0 \\ & A_{a}^{\prime a}=A_{a}^{a}-2 k \cdot C=A_{a}^{a}-2\left(C_{0}-C_{3}\right)=0 \tag{3.1953} \end{array}(3.1953)A00=A00+2C0C0+C3=A00+C0+C3=0(i3)A0i=A0i+C0ki+k0Ci=A0i+Ci=0Aaa=Aaa2kC=Aaa2(C0C3)=0
This is a linear set of equations for the four constants C a C a C^(a)C^{a}Ca with the solution
(3.1954) C 0 = 1 4 A a a 1 2 A 00 , C 3 = 1 4 A a a 1 2 A 00 , C 1 = A 01 , C 2 = A 02 . (3.1954) C 0 = 1 4 A a a 1 2 A 00 , C 3 = 1 4 A a a 1 2 A 00 , C 1 = A 01 , C 2 = A 02 . {:(3.1954)C_(0)=(1)/(4)A_(a)^(a)-(1)/(2)A_(00)","quadC_(3)=-(1)/(4)A_(a)^(a)-(1)/(2)A_(00)","quadC_(1)=-A_(01)","quadC_(2)=-A_(02).:}\begin{equation*} C_{0}=\frac{1}{4} A_{a}^{a}-\frac{1}{2} A_{00}, \quad C_{3}=-\frac{1}{4} A_{a}^{a}-\frac{1}{2} A_{00}, \quad C_{1}=-A_{01}, \quad C_{2}=-A_{02} . \tag{3.1954} \end{equation*}(3.1954)C0=14Aaa12A00,C3=14Aaa12A00,C1=A01,C2=A02.

2.133

a) Consider the quantities h μ ν h μ ν h_(mu nu)h_{\mu \nu}hμν to be perturbations or deviations of the components of the metric tensor away from flat spacetime. In general, for h μ ν h μ ν h_(mu nu)h_{\mu \nu}hμν, the Lorenz gauge condition is given by
(3.1955) v h μ v 1 2 μ h ν v = 0 . (3.1955) v h μ v 1 2 μ h ν v = 0 . {:(3.1955)del_(v)h_(mu)^(v)-(1)/(2)del_(mu)h_(nu)^(v)=0.:}\begin{equation*} \partial_{v} h_{\mu}^{v}-\frac{1}{2} \partial_{\mu} h_{\nu}^{v}=0 . \tag{3.1955} \end{equation*}(3.1955)vhμv12μhνv=0.
Cf. the analogy to this gauge condition in electromagnetism. If the Lorenz gauge condition is fulfilled, then Einstein's equations in vacuum reduce to h ¯ μ ν = 0 h ¯ μ ν = 0 ◻ bar(h)_(mu nu)=0\square \bar{h}_{\mu \nu}=0h¯μν=0, where h ¯ μ ν = h μ ν η μ ν h / 2 h ¯ μ ν = h μ ν η μ ν h / 2 bar(h)_(mu nu)=h_(mu nu)-eta_(mu nu)h//2\bar{h}_{\mu \nu}=h_{\mu \nu}-\eta_{\mu \nu} h / 2h¯μν=hμνημνh/2 with h = η μ ν h μ ν h = η μ ν h μ ν h=eta^(mu nu)h_(mu nu)h=\eta^{\mu \nu} h_{\mu \nu}h=ημνhμν. This gauge condition does not completely fix the coordinates, so in order to do so, further conditions can be imposed, e.g., h i 0 = 0 h i 0 = 0 h_(i0)=0h_{i 0}=0hi0=0 and η μ ν h μ ν = 0 η μ ν h μ ν = 0 eta^(mu nu)h_(mu nu)=0\eta^{\mu \nu} h_{\mu \nu}=0ημνhμν=0, which are collectively referred to as the transverse traceless gauge (or TT gauge), where h μ ν = h ¯ μ ν h μ ν = h ¯ μ ν h_(mu nu)= bar(h)_(mu nu)h_{\mu \nu}=\bar{h}_{\mu \nu}hμν=h¯μν. In particular, in the TT gauge, the Lorenz gauge condition reduces to ν h ν μ = 0 ν h ν μ = 0 del_(nu)h^(nu)_(mu)=0\partial_{\nu} h^{\nu}{ }_{\mu}=0νhνμ=0. The symmetric tensor h μ ν h μ ν h_(mu nu)h_{\mu \nu}hμν has ten independent components and the TT gauge contains eight conditions, which in fact means that only two components of h μ ν h μ ν h_(mu nu)h_{\mu \nu}hμν are independent. The two independent components of h μ ν h μ ν h_(mu nu)h_{\mu \nu}hμν are normally denoted h + h + h_(+)h_{+}h+and h × h × h_(xx)h_{\times}h×, which correspond to the two independent polarization states of a gravitational wave.
b) First, let us consider two particles, which are influenced by a plus-polarized gravitational wave along the z z zzz-direction such that
(3.1956) ( h μ v ) = diag ( 0 , h + , h + , 0 ) , h + = h 0 sin [ 2 π f ( t z ) ] , h × = 0 , (3.1956) h μ v = diag 0 , h + , h + , 0 , h + = h 0 sin [ 2 π f ( t z ) ] , h × = 0 , {:(3.1956)(h_(mu v))=diag(0,h_(+),-h_(+),0)","quadh_(+)=h_(0)sin[2pi f(t-z)]","quadh_(xx)=0",":}\begin{equation*} \left(h_{\mu v}\right)=\operatorname{diag}\left(0, h_{+},-h_{+}, 0\right), \quad h_{+}=h_{0} \sin [2 \pi f(t-z)], \quad h_{\times}=0, \tag{3.1956} \end{equation*}(3.1956)(hμv)=diag(0,h+,h+,0),h+=h0sin[2πf(tz)],h×=0,
where h 0 1 h 0 1 h_(0)≪1h_{0} \ll 1h01 is the amplitude and f f fff is the frequency.
Then, in fact, the geodesic equation for the displacement vector S μ S μ S_(mu)S_{\mu}Sμ is given by
(3.1957) d 2 S μ d t 2 = 1 2 d 2 h μ v d t 2 S v (3.1957) d 2 S μ d t 2 = 1 2 d 2 h μ v d t 2 S v {:(3.1957)(d^(2)S_(mu))/(dt^(2))=(1)/(2)(d^(2)h_(mu v))/(dt^(2))S^(v):}\begin{equation*} \frac{d^{2} S_{\mu}}{d t^{2}}=\frac{1}{2} \frac{d^{2} h_{\mu v}}{d t^{2}} S^{v} \tag{3.1957} \end{equation*}(3.1957)d2Sμdt2=12d2hμvdt2Sv
where h μ ν h μ ν h_(mu nu)h_{\mu \nu}hμν is the given gravitational wave. We observe that this gravitational wave affects neither S 0 S 0 S_(0)S_{0}S0 nor S 3 S 3 S_(3)S_{3}S3. Thus, the only effect on the geodesics is taking place in the x x xxx - and y y yyy-directions. Without loss of generality, we can therefore assume that z = 0 z = 0 z=0z=0z=0. In this case, the gravitational wave is plus-polarized only, i.e., h × = 0 h × = 0 h_(xx)=0h_{\times}=0h×=0, so that the geodesic equation simplifies to two equations for S 1 = S 1 S 1 = S 1 S_(1)=-S^(1)S_{1}=-S^{1}S1=S1 and S 2 = S 2 S 2 = S 2 S_(2)=-S^(2)S_{2}=-S^{2}S2=S2, namely
(3.1958) d 2 S 1 d t 2 = ( 2 π f ) 2 2 h 0 sin ( 2 π f t ) S 1 , d 2 S 2 d t 2 = ( 2 π f ) 2 2 h 0 sin ( 2 π f t ) S 2 (3.1958) d 2 S 1 d t 2 = ( 2 π f ) 2 2 h 0 sin ( 2 π f t ) S 1 , d 2 S 2 d t 2 = ( 2 π f ) 2 2 h 0 sin ( 2 π f t ) S 2 {:(3.1958)(d^(2)S_(1))/(dt^(2))=((2pi f)^(2))/(2)h_(0)sin(2pi ft)S_(1)","quad(d^(2)S_(2))/(dt^(2))=-((2pi f)^(2))/(2)h_(0)sin(2pi ft)S_(2):}\begin{equation*} \frac{d^{2} S_{1}}{d t^{2}}=\frac{(2 \pi f)^{2}}{2} h_{0} \sin (2 \pi f t) S_{1}, \quad \frac{d^{2} S_{2}}{d t^{2}}=-\frac{(2 \pi f)^{2}}{2} h_{0} \sin (2 \pi f t) S_{2} \tag{3.1958} \end{equation*}(3.1958)d2S1dt2=(2πf)22h0sin(2πft)S1,d2S2dt2=(2πf)22h0sin(2πft)S2
which can be solved perturbatively in h 0 h 0 h_(0)h_{0}h0. Up to first order in h 0 h 0 h_(0)h_{0}h0, we obtain
(3.1959) S 1 ( t ) = S 1 ( 0 ) [ 1 1 2 h 0 sin ( 2 π f t ) + ] (3.1960) S 2 ( t ) = S 2 ( 0 ) [ 1 + 1 2 h 0 sin ( 2 π f t ) + ] (3.1959) S 1 ( t ) = S 1 ( 0 ) 1 1 2 h 0 sin ( 2 π f t ) + (3.1960) S 2 ( t ) = S 2 ( 0 ) 1 + 1 2 h 0 sin ( 2 π f t ) + {:[(3.1959)S_(1)(t)=S_(1)(0)[1-(1)/(2)h_(0)sin(2pi ft)+cdots]],[(3.1960)S_(2)(t)=S_(2)(0)[1+(1)/(2)h_(0)sin(2pi ft)+cdots]]:}\begin{align*} & S_{1}(t)=S_{1}(0)\left[1-\frac{1}{2} h_{0} \sin (2 \pi f t)+\cdots\right] \tag{3.1959}\\ & S_{2}(t)=S_{2}(0)\left[1+\frac{1}{2} h_{0} \sin (2 \pi f t)+\cdots\right] \tag{3.1960} \end{align*}(3.1959)S1(t)=S1(0)[112h0sin(2πft)+](3.1960)S2(t)=S2(0)[1+12h0sin(2πft)+]
Next, the measured distance Δ x S 1 ( t ) Δ x S 1 ( t ) Delta x-=S_(1)(t)\Delta x \equiv S_{1}(t)ΔxS1(t) between the two particles, which was initially the distance Δ x 0 S 1 ( 0 ) Δ x 0 S 1 ( 0 ) Deltax_(0)-=S_(1)(0)\Delta x_{0} \equiv S_{1}(0)Δx0S1(0) along the x x xxx-direction, will be
(3.1961) Δ x Δ x 0 = S 1 ( t ) S 1 ( 0 ) 1 1 2 h 0 sin ( 2 π f t ) (3.1961) Δ x Δ x 0 = S 1 ( t ) S 1 ( 0 ) 1 1 2 h 0 sin ( 2 π f t ) {:(3.1961)(Delta x)/(Deltax_(0))=(S_(1)(t))/(S_(1)(0))≃1-(1)/(2)h_(0)sin(2pi ft):}\begin{equation*} \frac{\Delta x}{\Delta x_{0}}=\frac{S_{1}(t)}{S_{1}(0)} \simeq 1-\frac{1}{2} h_{0} \sin (2 \pi f t) \tag{3.1961} \end{equation*}(3.1961)ΔxΔx0=S1(t)S1(0)112h0sin(2πft)
which means that the relative distance δ x Δ x Δ x 0 δ x Δ x Δ x 0 delta x-=Delta x-Deltax_(0)\delta x \equiv \Delta x-\Delta x_{0}δxΔxΔx0 between the two particles oscillate with f f fff. This does not mean that the positions of the particle coordinates change, but the coordinates themselves oscillate.
Finally, assuming d d ddd to be the measured distance Δ x Δ x Delta x\Delta xΔx and L L LLL the initial distance Δ x 0 Δ x 0 Deltax_(0)\Delta x_{0}Δx0 between the two particles, we obtain
(3.1962) d L = 1 1 2 h 0 sin ( 2 π f t ) d = [ 1 1 2 h 0 sin ( 2 π f t ) ] L (3.1962) d L = 1 1 2 h 0 sin ( 2 π f t ) d = 1 1 2 h 0 sin ( 2 π f t ) L {:(3.1962)(d)/(L)=1-(1)/(2)*h_(0)sin(2pi ft)quad=>quad d=[1-(1)/(2)h_(0)sin(2pi ft)]L:}\begin{equation*} \frac{d}{L}=1-\frac{1}{2} \cdot h_{0} \sin (2 \pi f t) \quad \Rightarrow \quad d=\left[1-\frac{1}{2} h_{0} \sin (2 \pi f t)\right] L \tag{3.1962} \end{equation*}(3.1962)dL=112h0sin(2πft)d=[112h0sin(2πft)]L
which is what we wanted to show.

2.134

First, consider the energy-momentum tensor T μ ν T μ ν T^(mu nu)T^{\mu \nu}Tμν. Using T μ ν T μ ν T^(mu nu)T^{\mu \nu}Tμν and its conservation law, i.e., ν T μ ν = 0 ν T μ ν = 0 grad_(nu)T^(mu nu)=0\nabla_{\nu} T^{\mu \nu}=0νTμν=0, we can write
(3.1963) T μ ν = λ ( T μ λ x ν ) ( λ T μ λ ) x ν (3.1963) T μ ν = λ T μ λ x ν λ T μ λ x ν {:(3.1963)T^(mu nu)=del_(lambda)(T^(mu lambda)x^(nu))-(del_(lambda)T^(mu lambda))x^(nu):}\begin{equation*} T^{\mu \nu}=\partial_{\lambda}\left(T^{\mu \lambda} x^{\nu}\right)-\left(\partial_{\lambda} T^{\mu \lambda}\right) x^{\nu} \tag{3.1963} \end{equation*}(3.1963)Tμν=λ(Tμλxν)(λTμλ)xν
Note that in the linearized approximation, v T μ ν = 0 v T μ ν = 0 grad_(v)T^(mu nu)=0\nabla_{v} T^{\mu \nu}=0vTμν=0 reduces to ν T μ ν = 0 ν T μ ν = 0 del_(nu)T^(mu nu)=0\partial_{\nu} T^{\mu \nu}=0νTμν=0. For the spatial components of T μ ν T μ ν T^(mu nu)T^{\mu \nu}Tμν, using the fact that 0 ( T i 0 x j ) ( 0 T i 0 ) x j = 0 0 T i 0 x j 0 T i 0 x j = 0 del_(0)(T^(i0)x^(j))-(del_(0)T^(i0))x^(j)=0\partial_{0}\left(T^{i 0} x^{j}\right)-\left(\partial_{0} T^{i 0}\right) x^{j}=00(Ti0xj)(0Ti0)xj=0, we have
(3.1964) T i j = k ( T i k x j ) ( k T i k ) x j = k ( T i k x j ) [ ( v T i v ) x j ( 0 T i 0 ) x j ] (3.1964) T i j = k T i k x j k T i k x j = k T i k x j v T i v x j 0 T i 0 x j {:(3.1964)T^(ij)=del_(k)(T^(ik)x^(j))-(del_(k)T^(ik))x^(j)=del_(k)(T^(ik)x^(j))-[(del_(v)T^(iv))x^(j)-(del_(0)T^(i0))x^(j)]:}\begin{equation*} T^{i j}=\partial_{k}\left(T^{i k} x^{j}\right)-\left(\partial_{k} T^{i k}\right) x^{j}=\partial_{k}\left(T^{i k} x^{j}\right)-\left[\left(\partial_{v} T^{i v}\right) x^{j}-\left(\partial_{0} T^{i 0}\right) x^{j}\right] \tag{3.1964} \end{equation*}(3.1964)Tij=k(Tikxj)(kTik)xj=k(Tikxj)[(vTiv)xj(0Ti0)xj]
since v T i v = 0 T i 0 + k T i k = 0 v T i v = 0 T i 0 + k T i k = 0 del_(v)T^(iv)=del_(0)T^(i0)+del_(k)T^(ik)=0\partial_{v} T^{i v}=\partial_{0} T^{i 0}+\partial_{k} T^{i k}=0vTiv=0Ti0+kTik=0, and therefore, we find that
(3.1965) T i j = k ( T i k x j ) + ( 0 T i 0 ) x j = k ( T i k x j ) + 0 ( T i 0 x j ) (3.1965) T i j = k T i k x j + 0 T i 0 x j = k T i k x j + 0 T i 0 x j {:(3.1965)T^(ij)=del_(k)(T^(ik)x^(j))+(del_(0)T^(i0))x^(j)=del_(k)(T^(ik)x^(j))+del_(0)(T^(i0)x^(j)):}\begin{equation*} T^{i j}=\partial_{k}\left(T^{i k} x^{j}\right)+\left(\partial_{0} T^{i 0}\right) x^{j}=\partial_{k}\left(T^{i k} x^{j}\right)+\partial_{0}\left(T^{i 0} x^{j}\right) \tag{3.1965} \end{equation*}(3.1965)Tij=k(Tikxj)+(0Ti0)xj=k(Tikxj)+0(Ti0xj)
where we used 0 ( T i 0 x j ) = ( 0 T i 0 ) x j + T i 0 0 x j = ( 0 T i 0 ) x j 0 T i 0 x j = 0 T i 0 x j + T i 0 0 x j = 0 T i 0 x j del_(0)(T^(i0)x^(j))=(del_(0)T^(i0))x^(j)+T^(i0)del_(0)x^(j)=(del_(0)T^(i0))x^(j)\partial_{0}\left(T^{i 0} x^{j}\right)=\left(\partial_{0} T^{i 0}\right) x^{j}+T^{i 0} \partial_{0} x^{j}=\left(\partial_{0} T^{i 0}\right) x^{j}0(Ti0xj)=(0Ti0)xj+Ti00xj=(0Ti0)xj. Similarly, for the T i 0 x j T i 0 x j T^(i0)x^(j)T^{i 0} x^{j}Ti0xj term, we use the same rewriting technique, i.e.,
T i 0 x j = k ( T k 0 x j x i ) k ( T k 0 x j ) x i = k ( T k 0 x j x i ) ( k T k 0 ) x j x i T k 0 ( k x j ) x i (3.1966) = k ( T k 0 x j x i ) ( k T k 0 ) x j x i T j 0 x i T i 0 x j = k T k 0 x j x i k T k 0 x j x i = k T k 0 x j x i k T k 0 x j x i T k 0 k x j x i (3.1966) = k T k 0 x j x i k T k 0 x j x i T j 0 x i {:[T^(i0)x^(j)=del_(k)(T^(k0)x^(j)x^(i))-del_(k)(T^(k0)x^(j))x^(i)=del_(k)(T^(k0)x^(j)x^(i))-(del_(k)T^(k0))x^(j)x^(i)-T^(k0)(del_(k)x^(j))x^(i)],[(3.1966)=del_(k)(T^(k0)x^(j)x^(i))-(del_(k)T^(k0))x^(j)x^(i)-T^(j0)x^(i)]:}\begin{align*} T^{i 0} x^{j} & =\partial_{k}\left(T^{k 0} x^{j} x^{i}\right)-\partial_{k}\left(T^{k 0} x^{j}\right) x^{i}=\partial_{k}\left(T^{k 0} x^{j} x^{i}\right)-\left(\partial_{k} T^{k 0}\right) x^{j} x^{i}-T^{k 0}\left(\partial_{k} x^{j}\right) x^{i} \\ & =\partial_{k}\left(T^{k 0} x^{j} x^{i}\right)-\left(\partial_{k} T^{k 0}\right) x^{j} x^{i}-T^{j 0} x^{i} \tag{3.1966} \end{align*}Ti0xj=k(Tk0xjxi)k(Tk0xj)xi=k(Tk0xjxi)(kTk0)xjxiTk0(kxj)xi(3.1966)=k(Tk0xjxi)(kTk0)xjxiTj0xi
Now, symmetrizing the tensor structure T i 0 x j T i 0 x j T^(i0)x^(j)T^{i 0} x^{j}Ti0xj, we have
T i 0 x j ^ 1 2 ( T i 0 x j + T j 0 x i ) = 1 2 [ k ( T k 0 x j x i ) + k ( T k 0 x i x j ) ( k T k 0 ) x j x i ( k T k 0 ) x i x j T j 0 x i T i 0 x j ] = k ( T k 0 x i x j ) ( k T k 0 ) x i x j 1 2 ( T i 0 x j + T j 0 x i ) (3.1967) = k ( T k 0 x i x j ) ( k T k 0 ) x i x j T i 0 x j ^ T i 0 x j ^ 1 2 T i 0 x j + T j 0 x i = 1 2 k T k 0 x j x i + k T k 0 x i x j k T k 0 x j x i k T k 0 x i x j T j 0 x i T i 0 x j = k T k 0 x i x j k T k 0 x i x j 1 2 T i 0 x j + T j 0 x i (3.1967) = k T k 0 x i x j k T k 0 x i x j T i 0 x j ^ {:[ widehat(T^(i0)x^(j))-=(1)/(2)(T^(i0)x^(j)+T^(j0)x^(i))],[=(1)/(2)[del_(k)(T^(k0)x^(j)x^(i))+del_(k)(T^(k0)x^(i)x^(j))-(del_(k)T^(k0))x^(j)x^(i):}],[{:-(del_(k)T^(k0))x^(i)x^(j)-T^(j0)x^(i)-T^(i0)x^(j)]],[=del_(k)(T^(k0)x^(i)x^(j))-(del_(k)T^(k0))x^(i)x^(j)-(1)/(2)(T^(i0)x^(j)+T^(j0)x^(i))],[(3.1967)=del_(k)(T^(k0)x^(i)x^(j))-(del_(k)T^(k0))x^(i)x^(j)- widehat(T^(i0)x^(j))]:}\begin{align*} \widehat{T^{i 0} x^{j}} \equiv & \frac{1}{2}\left(T^{i 0} x^{j}+T^{j 0} x^{i}\right) \\ = & \frac{1}{2}\left[\partial_{k}\left(T^{k 0} x^{j} x^{i}\right)+\partial_{k}\left(T^{k 0} x^{i} x^{j}\right)-\left(\partial_{k} T^{k 0}\right) x^{j} x^{i}\right. \\ & \left.-\left(\partial_{k} T^{k 0}\right) x^{i} x^{j}-T^{j 0} x^{i}-T^{i 0} x^{j}\right] \\ = & \partial_{k}\left(T^{k 0} x^{i} x^{j}\right)-\left(\partial_{k} T^{k 0}\right) x^{i} x^{j}-\frac{1}{2}\left(T^{i 0} x^{j}+T^{j 0} x^{i}\right) \\ = & \partial_{k}\left(T^{k 0} x^{i} x^{j}\right)-\left(\partial_{k} T^{k 0}\right) x^{i} x^{j}-\widehat{T^{i 0} x^{j}} \tag{3.1967} \end{align*}Ti0xj^12(Ti0xj+Tj0xi)=12[k(Tk0xjxi)+k(Tk0xixj)(kTk0)xjxi(kTk0)xixjTj0xiTi0xj]=k(Tk0xixj)(kTk0)xixj12(Ti0xj+Tj0xi)(3.1967)=k(Tk0xixj)(kTk0)xixjTi0xj^
which leads to
(3.1968) T i 0 x j ^ = 1 2 k ( T k 0 x i x j ) 1 2 ( k T k 0 ) x i x j (3.1968) T i 0 x j ^ = 1 2 k T k 0 x i x j 1 2 k T k 0 x i x j {:(3.1968) widehat(T^(i0)x^(j))=(1)/(2)del_(k)(T^(k0)x^(i)x^(j))-(1)/(2)(del_(k)T^(k0))x^(i)x^(j):}\begin{equation*} \widehat{T^{i 0} x^{j}}=\frac{1}{2} \partial_{k}\left(T^{k 0} x^{i} x^{j}\right)-\frac{1}{2}\left(\partial_{k} T^{k 0}\right) x^{i} x^{j} \tag{3.1968} \end{equation*}(3.1968)Ti0xj^=12k(Tk0xixj)12(kTk0)xixj
Again, using v T ν 0 = 0 T 00 + k T k 0 = 0 v T ν 0 = 0 T 00 + k T k 0 = 0 del_(v)T^(nu0)=del_(0)T^(00)+del_(k)T^(k0)=0\partial_{v} T^{\nu 0}=\partial_{0} T^{00}+\partial_{k} T^{k 0}=0vTν0=0T00+kTk0=0, we find that
(3.1969) T i 0 x j ^ = 1 2 k ( T k 0 x i x j ) + 1 2 ( 0 T 00 ) x i x j (3.1969) T i 0 x j ^ = 1 2 k T k 0 x i x j + 1 2 0 T 00 x i x j {:(3.1969) widehat(T^(i0)x^(j))=(1)/(2)del_(k)(T^(k0)x^(i)x^(j))+(1)/(2)(del_(0)T^(00))x^(i)x^(j):}\begin{equation*} \widehat{T^{i 0} x^{j}}=\frac{1}{2} \partial_{k}\left(T^{k 0} x^{i} x^{j}\right)+\frac{1}{2}\left(\partial_{0} T^{00}\right) x^{i} x^{j} \tag{3.1969} \end{equation*}(3.1969)Ti0xj^=12k(Tk0xixj)+12(0T00)xixj
Since the energy-momentum tensor is symmetric, i.e., T μ ν = T ν μ T μ ν = T ν μ T^(mu nu)=T^(nu mu)T^{\mu \nu}=T^{\nu \mu}Tμν=Tνμ, we can then combine the expressions for T i j T i j T^(ij)T^{i j}Tij and T i 0 x j ^ T i 0 x j ^ widehat(T^(i0)x^(j))\widehat{T^{i 0} x^{j}}Ti0xj^, and thus, we obtain
T i j = T i j ^ 1 2 ( T i j + T j i ) = 1 2 [ k ( T i k x j ) + k ( T j k x i ) ] + 1 2 [ 0 ( T i 0 x j ) + 0 ( T j 0 x i ) ] T i j = T i j ^ 1 2 T i j + T j i = 1 2 k T i k x j + k T j k x i + 1 2 0 T i 0 x j + 0 T j 0 x i {:[T^(ij)= widehat(T^(ij))-=(1)/(2)(T^(ij)+T^(ji))],[=(1)/(2)[del_(k)(T^(ik)x^(j))+del_(k)(T^(jk)x^(i))]+(1)/(2)[del_(0)(T^(i0)x^(j))+del_(0)(T^(j0)x^(i))]]:}\begin{aligned} T^{i j} & =\widehat{T^{i j}} \equiv \frac{1}{2}\left(T^{i j}+T^{j i}\right) \\ & =\frac{1}{2}\left[\partial_{k}\left(T^{i k} x^{j}\right)+\partial_{k}\left(T^{j k} x^{i}\right)\right]+\frac{1}{2}\left[\partial_{0}\left(T^{i 0} x^{j}\right)+\partial_{0}\left(T^{j 0} x^{i}\right)\right] \end{aligned}Tij=Tij^12(Tij+Tji)=12[k(Tikxj)+k(Tjkxi)]+12[0(Ti0xj)+0(Tj0xi)]
= 1 2 k ( T i k x j + T j k x i ) + 0 T i 0 x j ^ (3.1970) = 1 2 k ( T i k x j + T j k x i ) + 1 2 0 k ( T k 0 x i x j ) + 1 2 0 [ ( 0 T 00 ) x i x j ] = 1 2 k T i k x j + T j k x i + 0 T i 0 x j ^ (3.1970) = 1 2 k T i k x j + T j k x i + 1 2 0 k T k 0 x i x j + 1 2 0 0 T 00 x i x j {:[=(1)/(2)del_(k)(T^(ik)x^(j)+T^(jk)x^(i))+del_(0) widehat(T^(i0)x^(j))],[(3.1970)=(1)/(2)*del_(k)(T^(ik)x^(j)+T^(jk)x^(i))+(1)/(2)*del_(0)del_(k)(T^(k0)x^(i)x^(j))+(1)/(2)del_(0)[(del_(0)T^(00))x^(i)x^(j)]]:}\begin{align*} & =\frac{1}{2} \partial_{k}\left(T^{i k} x^{j}+T^{j k} x^{i}\right)+\partial_{0} \widehat{T^{i 0} x^{j}} \\ & =\frac{1}{2} \cdot \partial_{k}\left(T^{i k} x^{j}+T^{j k} x^{i}\right)+\frac{1}{2} \cdot \partial_{0} \partial_{k}\left(T^{k 0} x^{i} x^{j}\right)+\frac{1}{2} \partial_{0}\left[\left(\partial_{0} T^{00}\right) x^{i} x^{j}\right] \tag{3.1970} \end{align*}=12k(Tikxj+Tjkxi)+0Ti0xj^(3.1970)=12k(Tikxj+Tjkxi)+120k(Tk0xixj)+120[(0T00)xixj]
In addition, using that 0 k = k 0 0 k = k 0 del_(0)del_(k)=del_(k)del_(0)\partial_{0} \partial_{k}=\partial_{k} \partial_{0}0k=k0 and 0 [ ( 0 T 00 ) x i x j ] = ( 0 2 T 00 ) x i x j 0 0 T 00 x i x j = 0 2 T 00 x i x j del_(0)[(del_(0)T^(00))x^(i)x^(j)]=(del_(0)^(2)T^(00))x^(i)x^(j)\partial_{0}\left[\left(\partial_{0} T^{00}\right) x^{i} x^{j}\right]=\left(\partial_{0}{ }^{2} T^{00}\right) x^{i} x^{j}0[(0T00)xixj]=(02T00)xixj, we have
T i j = 1 2 k ( T i k x j + T j k x i ) + 1 2 k 0 ( T k 0 x i x j ) + 1 2 ( 0 2 T 00 ) x i x j (3.1971) = k [ 1 2 ( T i k x j + T j k x i ) + 1 2 0 ( T k 0 x i x j ) ] + 1 2 ( 0 2 T 00 ) x i x j T i j = 1 2 k T i k x j + T j k x i + 1 2 k 0 T k 0 x i x j + 1 2 0 2 T 00 x i x j (3.1971) = k 1 2 T i k x j + T j k x i + 1 2 0 T k 0 x i x j + 1 2 0 2 T 00 x i x j {:[T^(ij)=(1)/(2)del_(k)(T^(ik)x^(j)+T^(jk)x^(i))+(1)/(2)del_(k)del_(0)(T^(k0)x^(i)x^(j))+(1)/(2)(del_(0)^(2)T^(00))x^(i)x^(j)],[(3.1971)=del_(k)[(1)/(2)(T^(ik)x^(j)+T^(jk)x^(i))+(1)/(2)*del_(0)(T^(k0)x^(i)x^(j))]+(1)/(2)(del_(0)^(2)T^(00))x^(i)x^(j)]:}\begin{align*} T^{i j} & =\frac{1}{2} \partial_{k}\left(T^{i k} x^{j}+T^{j k} x^{i}\right)+\frac{1}{2} \partial_{k} \partial_{0}\left(T^{k 0} x^{i} x^{j}\right)+\frac{1}{2}\left(\partial_{0}{ }^{2} T^{00}\right) x^{i} x^{j} \\ & =\partial_{k}\left[\frac{1}{2}\left(T^{i k} x^{j}+T^{j k} x^{i}\right)+\frac{1}{2} \cdot \partial_{0}\left(T^{k 0} x^{i} x^{j}\right)\right]+\frac{1}{2}\left(\partial_{0}{ }^{2} T^{00}\right) x^{i} x^{j} \tag{3.1971} \end{align*}Tij=12k(Tikxj+Tjkxi)+12k0(Tk0xixj)+12(02T00)xixj(3.1971)=k[12(Tikxj+Tjkxi)+120(Tk0xixj)]+12(02T00)xixj
Next, using the definition
(3.1972) h ¯ μ ν ( t , x ) 4 r T μ ν ( t r , x ) d 3 x (3.1972) h ¯ μ ν ( t , x ) 4 r T μ ν t r , x d 3 x {:(3.1972) bar(h)_(mu nu)(t","x)∼(4)/(r)intT_(mu nu)(t-r,x^('))d^(3)x^('):}\begin{equation*} \bar{h}_{\mu \nu}(t, \mathbf{x}) \sim \frac{4}{r} \int T_{\mu \nu}\left(t-r, \mathbf{x}^{\prime}\right) d^{3} x^{\prime} \tag{3.1972} \end{equation*}(3.1972)h¯μν(t,x)4rTμν(tr,x)d3x
and choosing the spatial components, we have
(3.1973) h ¯ i j 4 r T i j d 3 x (3.1973) h ¯ i j 4 r T i j d 3 x {:(3.1973) bar(h)_(ij)∼(4)/(r)intT_(ij)^(')d^(3)x^('):}\begin{equation*} \bar{h}_{i j} \sim \frac{4}{r} \int T_{i j}^{\prime} d^{3} x^{\prime} \tag{3.1973} \end{equation*}(3.1973)h¯ij4rTijd3x
Inserting the symmetrized version of the energy-momentum tensor, we obtain
(3.1974) h ¯ i j 4 r 1 2 ( 0 2 T 00 ) x i x j d 3 x (3.1974) h ¯ i j 4 r 1 2 0 2 T 00 x i x j d 3 x {:(3.1974) bar(h)_(ij)∼(4)/(r)int(1)/(2)(del_(0)^(2)T_(00)^('))x_(i)^(')x_(j)^(')d^(3)x^('):}\begin{equation*} \bar{h}_{i j} \sim \frac{4}{r} \int \frac{1}{2}\left(\partial_{0}^{2} T_{00}^{\prime}\right) x_{i}^{\prime} x_{j}^{\prime} d^{3} x^{\prime} \tag{3.1974} \end{equation*}(3.1974)h¯ij4r12(02T00)xixjd3x
since we can drop the terms with total derivatives with respect to the spatial components. Therefore, we find that
(3.1975) h ¯ i j 2 r ( 0 2 T 00 ) x i x j d 3 x (3.1975) h ¯ i j 2 r 0 2 T 00 x i x j d 3 x {:(3.1975) bar(h)_(ij)∼(2)/(r)int(del_(0)^(2)T_(00)^('))x_(i)^(')x_(j)^(')d^(3)x^('):}\begin{equation*} \bar{h}_{i j} \sim \frac{2}{r} \int\left(\partial_{0}^{2} T_{00}^{\prime}\right) x_{i}^{\prime} x_{j}^{\prime} d^{3} x^{\prime} \tag{3.1975} \end{equation*}(3.1975)h¯ij2r(02T00)xixjd3x
and using the fact that T 00 = T 00 = ρ T 00 = T 00 = ρ T_(00)=T^(00)=rhoT_{00}=T^{00}=\rhoT00=T00=ρ, we finally obtain
h ¯ i j 2 r 2 ρ t 2 x i x j d 3 x = 2 r d 2 d t 2 ρ x i x j d 3 x = 2 r d 2 d t 2 ρ ( t r , x ) x i x j d 3 x h ¯ i j 2 r 2 ρ t 2 x i x j d 3 x = 2 r d 2 d t 2 ρ x i x j d 3 x = 2 r d 2 d t 2 ρ t r , x x i x j d 3 x bar(h)_(ij)∼(2)/(r)int(del^(2)rho^('))/(delt^(2))x_(i)^(')x_(j)^(')d^(3)x^(')=(2)/(r)(d^(2))/(dt^(2))intrho^(')x_(i)^(')x_(j)^(')d^(3)x^(')=(2)/(r)(d^(2))/(dt^(2))int rho(t-r,x^('))x_(i)^(')x_(j)^(')d^(3)x^(')\bar{h}_{i j} \sim \frac{2}{r} \int \frac{\partial^{2} \rho^{\prime}}{\partial t^{2}} x_{i}^{\prime} x_{j}^{\prime} d^{3} x^{\prime}=\frac{2}{r} \frac{d^{2}}{d t^{2}} \int \rho^{\prime} x_{i}^{\prime} x_{j}^{\prime} d^{3} x^{\prime}=\frac{2}{r} \frac{d^{2}}{d t^{2}} \int \rho\left(t-r, \mathbf{x}^{\prime}\right) x_{i}^{\prime} x_{j}^{\prime} d^{3} x^{\prime}h¯ij2r2ρt2xixjd3x=2rd2dt2ρxixjd3x=2rd2dt2ρ(tr,x)xixjd3x,
where we used Leibniz's integral rule twice.

2.135

For the estimates in this problem, we use that the amplitude of the gravitational waves is approximated by the relation
(3.1977) h M 2 r R M 5 / 3 r ( 4 π P ) 2 / 3 (3.1977) h M 2 r R M 5 / 3 r 4 π P 2 / 3 {:(3.1977)h prop(M^(2))/(rR)∼(M^(5//3))/(r)((4pi)/(P))^(2//3):}\begin{equation*} h \propto \frac{M^{2}}{r R} \sim \frac{M^{5 / 3}}{r}\left(\frac{4 \pi}{P}\right)^{2 / 3} \tag{3.1977} \end{equation*}(3.1977)hM2rRM5/3r(4πP)2/3
where M M MMM is the mass of the binary, r r rrr is the distance to the binary, and P P PPP is the period of one orbit of the binary, i.e., the orbital period. In conventional units, the
amplitude of the gravitational waves can be written as [see, e.g., K.D. Kokkotas, Gravitational Waves, Acta Phys. Polon. B 38, 3891 (2007)]
(3.1978) h 5 10 22 ( M 2.8 M ) 5 / 3 ( f 100 Hz ) 2 / 3 ( 15 Mpc r ) (3.1978) h 5 10 22 M 2.8 M 5 / 3 f 100 Hz 2 / 3 15 Mpc r {:(3.1978)h∼5*10^(-22)((M)/(2.8M_(o.)))^(5//3)((f)/(100(Hz)))^(2//3)((15Mpc)/(r)):}\begin{equation*} h \sim 5 \cdot 10^{-22}\left(\frac{M}{2.8 M_{\odot}}\right)^{5 / 3}\left(\frac{f}{100 \mathrm{~Hz}}\right)^{2 / 3}\left(\frac{15 \mathrm{Mpc}}{r}\right) \tag{3.1978} \end{equation*}(3.1978)h51022(M2.8M)5/3(f100 Hz)2/3(15Mpcr)
where f 1 / P f 1 / P f-=1//Pf \equiv 1 / Pf1/P is the frequency.
a) Using M 2.8 M M 2.8 M M≃2.8M_(o.)M \simeq 2.8 M_{\odot}M2.8M (where M 1.988435 10 30 kg M 1.988435 10 30 kg M_(o.)≃1.988435*10^(30)kgM_{\odot} \simeq 1.988435 \cdot 10^{30} \mathrm{~kg}M1.9884351030 kg is the solar mass), r = 5 kpc r = 5 kpc r=5kpcr=5 \mathrm{kpc}r=5kpc, and P = 1 h P = 1 h P=1hP=1 \mathrm{~h}P=1 h (which means that f = 1 / 3600 Hz f = 1 / 3600 Hz f=1//3600Hzf=1 / 3600 \mathrm{~Hz}f=1/3600 Hz ), we find that
(3.1979) h 5 10 22 1 5 / 3 ( 1 3.6 10 5 ) 2 / 3 3 10 3 3 10 22 10 22 (3.1979) h 5 10 22 1 5 / 3 1 3.6 10 5 2 / 3 3 10 3 3 10 22 10 22 {:(3.1979)h∼5*10^(-22)*1^(5//3)*((1)/(3.6*10^(5)))^(2//3)*3*10^(3)≃3*10^(-22)∼10^(-22):}\begin{equation*} h \sim 5 \cdot 10^{-22} \cdot 1^{5 / 3} \cdot\left(\frac{1}{3.6 \cdot 10^{5}}\right)^{2 / 3} \cdot 3 \cdot 10^{3} \simeq 3 \cdot 10^{-22} \sim 10^{-22} \tag{3.1979} \end{equation*}(3.1979)h5102215/3(13.6105)2/33103310221022
b) Again, using M 2.8 M M 2.8 M M≃2.8M_(o.)M \simeq 2.8 M_{\odot}M2.8M, but r = 15 Mpc r = 15 Mpc r=15Mpcr=15 \mathrm{Mpc}r=15Mpc and P = 0.02 s P = 0.02 s P=0.02sP=0.02 \mathrm{~s}P=0.02 s (which means that f = 50 Hz f = 50 Hz f=50Hzf=50 \mathrm{~Hz}f=50 Hz ), we find that
(3.1980) h 5 10 22 ( 1 2 ) 2 / 3 1 3 10 22 10 22 (3.1980) h 5 10 22 1 2 2 / 3 1 3 10 22 10 22 {:(3.1980)h∼5*10^(-22)*((1)/(2))^(2//3)*1≃3*10^(-22)∼10^(-22):}\begin{equation*} h \sim 5 \cdot 10^{-22} \cdot\left(\frac{1}{2}\right)^{2 / 3} \cdot 1 \simeq 3 \cdot 10^{-22} \sim 10^{-22} \tag{3.1980} \end{equation*}(3.1980)h51022(12)2/31310221022
which is basically the same result as in a).
c) For a binary system, assuming circular binary orbits, Kepler's third law provides a direct and accurate estimate for the orbital separation distance between the two binaries, i.e.,
(3.1981) 2 G ( M 1 + M 2 ) = ( 2 π P ) 2 ( R 1 + R 2 ) 3 (3.1981) 2 G M 1 + M 2 = 2 π P 2 R 1 + R 2 3 {:(3.1981)2G(M_(1)+M_(2))=((2pi)/(P))^(2)(R_(1)+R_(2))^(3):}\begin{equation*} 2 G\left(M_{1}+M_{2}\right)=\left(\frac{2 \pi}{P}\right)^{2}\left(R_{1}+R_{2}\right)^{3} \tag{3.1981} \end{equation*}(3.1981)2G(M1+M2)=(2πP)2(R1+R2)3
where G 6.674 10 11 N m 2 / kg 2 G 6.674 10 11 N m 2 / kg 2 G≃6.674*10^(-11)Nm^(2)//kg^(2)G \simeq 6.674 \cdot 10^{-11} \mathrm{~N} \mathrm{~m}^{2} / \mathrm{kg}^{2}G6.6741011 N m2/kg2 is Newton's gravitational constant, M 1 M 1 M_(1)M_{1}M1 and M 2 M 2 M_(2)M_{2}M2 are the masses of the two binaries, respectively, R 1 R 1 R_(1)R_{1}R1 and R 2 R 2 R_(2)R_{2}R2 are the respective distances to their common orbital center, and P P PPP is the orbital period. Assuming M 1 = M 2 M / 2 M 1 = M 2 M / 2 M_(1)=M_(2)-=M//2M_{1}=M_{2} \equiv M / 2M1=M2M/2 and R 1 = R 2 R R 1 = R 2 R R_(1)=R_(2)-=RR_{1}=R_{2} \equiv RR1=R2R, i.e., the binaries are equally heavy and they are at the same distance compared to their orbital center, we obtain (half of) the orbital separation distance R R RRR between the two binaries as
(3.1982) R = 1 2 G M P 2 4 π 2 3 (3.1982) R = 1 2 G M P 2 4 π 2 3 {:(3.1982)R=(1)/(2)root(3)((GMP^(2))/(4pi^(2))):}\begin{equation*} R=\frac{1}{2} \sqrt[3]{\frac{G M P^{2}}{4 \pi^{2}}} \tag{3.1982} \end{equation*}(3.1982)R=12GMP24π23
Inserting M = 2.8 M M = 2.8 M M=2.8M_(o.)M=2.8 M_{\odot}M=2.8M and P = 0.02 s P = 0.02 s P=0.02sP=0.02 \mathrm{~s}P=0.02 s, we find that
(3.1983) R 155600 m 100 km (3.1983) R 155600 m 100 km {:(3.1983)R≃155600m∼100km:}\begin{equation*} R \simeq 155600 \mathrm{~m} \sim 100 \mathrm{~km} \tag{3.1983} \end{equation*}(3.1983)R155600 m100 km
Thus, we can only hope to detect inspirals of compact binary systems (e.g., NS-NS, NS-BH, or BH-BH) with Earth-based interferometers such as LIGO.
d) For a spinning neutron star, the amplitude of the gravitational waves h h hhh is approximately given by
(3.1984) h 2 δ M R 2 Ω 2 r (3.1984) h 2 δ M R 2 Ω 2 r {:(3.1984)h prop(2delta MR^(2)Omega^(2))/(r):}\begin{equation*} h \propto \frac{2 \delta M R^{2} \Omega^{2}}{r} \tag{3.1984} \end{equation*}(3.1984)h2δMR2Ω2r
where δ M δ M delta M\delta MδM is the mass of a nonspherical deformation on the equator of the neutron star, R R RRR is the radius of the neutron star, Ω 2 π / P Ω 2 π / P Omega-=2pi//P\Omega \equiv 2 \pi / PΩ2π/P is the angular velocity expressed in the spin period P P PPP, and r r rrr is the distance to the neutron star. In conventional units, the amplitude of the gravitational waves can be written as [see, e.g., R. Prix, Gravitational Waves from Spinning Neutron Stars, in: W. Becker (ed.), Neutron Stars and Pulsars, Astrophysics and Space Science Library 357, 651-685, Springer (Berlin, 2009)]
(3.1985) h 10 2 G c 4 δ M R 2 f 2 r 3 10 25 ( δ M R 2 10 32 kg m 2 ) ( f 100 Hz ) 2 ( 100 pc r ) (3.1985) h 10 2 G c 4 δ M R 2 f 2 r 3 10 25 δ M R 2 10 32 kg m 2 f 100 Hz 2 100 pc r {:(3.1985)h prop10^(2)(G)/(c^(4))*(delta MR^(2)f^(2))/(r)∼3*10^(-25)((delta MR^(2))/(10^(32)(kg)m^(2)))((f)/(100(Hz)))^(2)((100pc)/(r)):}\begin{equation*} h \propto 10^{2} \frac{G}{c^{4}} \cdot \frac{\delta M R^{2} f^{2}}{r} \sim 3 \cdot 10^{-25}\left(\frac{\delta M R^{2}}{10^{32} \mathrm{~kg} \mathrm{~m}^{2}}\right)\left(\frac{f}{100 \mathrm{~Hz}}\right)^{2}\left(\frac{100 \mathrm{pc}}{r}\right) \tag{3.1985} \end{equation*}(3.1985)h102Gc4δMR2f2r31025(δMR21032 kg m2)(f100 Hz)2(100pcr)
where G G GGG is again Newton's gravitational constant, c c ccc is the speed of light in vacuum, and f f fff is the spin frequency. Inserting δ M = 10 6 M , R = 10 km , f = 50 Hz δ M = 10 6 M , R = 10 km , f = 50 Hz delta M=10^(-6)M_(o.),R=10km,f=50Hz\delta M=10^{-6} M_{\odot}, R=10 \mathrm{~km}, f=50 \mathrm{~Hz}δM=106M,R=10 km,f=50 Hz, and r = 1 kpc r = 1 kpc r=1kpcr=1 \mathrm{kpc}r=1kpc, we find that
(3.1986) h 3 10 25 2 ( 1 2 ) 2 1 10 = 1.5 10 26 10 26 (3.1986) h 3 10 25 2 1 2 2 1 10 = 1.5 10 26 10 26 {:(3.1986)h∼3*10^(-25)*2*((1)/(2))^(2)*(1)/(10)=1.5*10^(-26)∼10^(-26):}\begin{equation*} h \sim 3 \cdot 10^{-25} \cdot 2 \cdot\left(\frac{1}{2}\right)^{2} \cdot \frac{1}{10}=1.5 \cdot 10^{-26} \sim 10^{-26} \tag{3.1986} \end{equation*}(3.1986)h310252(12)2110=1.510261026
Thus, for the spinning neutron star, the order of magnitude of the amplitude of the gravitational waves at Earth is 10 26 10 26 10^(-26)10^{-26}1026.

2.136

a) The energy flux (or energy density) in units of power/area of gravitational waves can be estimated by the relation
(3.1987) F c 3 32 π 2 G | h ˙ | 2 c 3 8 G h 2 f 2 (3.1987) F c 3 32 π 2 G | h ˙ | 2 c 3 8 G h 2 f 2 {:(3.1987)F prop(c^(3))/(32pi^(2)G)|h^(˙)|^(2)∼(c^(3))/(8G)h^(2)f^(2):}\begin{equation*} F \propto \frac{c^{3}}{32 \pi^{2} G}|\dot{h}|^{2} \sim \frac{c^{3}}{8 G} h^{2} f^{2} \tag{3.1987} \end{equation*}(3.1987)Fc332π2G|h˙|2c38Gh2f2
where h h hhh and f f fff are the amplitude and the frequency, respectively, of the gravitational waves, which are assumed to be monochromatic. Note that the components of the metric tensor play the role of the gravitational potential, so the derivatives of the components of the metric tensor act as the field, and therefore, F | h ˙ | 2 F | h ˙ | 2 F prop|h^(˙)|^(2)F \propto|\dot{h}|^{2}F|h˙|2. A useful formula for an estimate of the energy flux on Earth due to gravitational waves is given by [see, e.g., I. Ciufolini (ed.) et al., Gravitational Waves, IoP (Bristol, 2001)]
(3.1988) F 3 10 3 ( h 10 22 ) 2 ( f 1 kHz ) 2 W / m 2 (3.1988) F 3 10 3 h 10 22 2 f 1 kHz 2 W / m 2 {:(3.1988)F∼3*10^(-3)((h)/(10^(-22)))^(2)((f)/(1kHz))^(2)W//m^(2):}\begin{equation*} F \sim 3 \cdot 10^{-3}\left(\frac{h}{10^{-22}}\right)^{2}\left(\frac{f}{1 \mathrm{kHz}}\right)^{2} \mathrm{~W} / \mathrm{m}^{2} \tag{3.1988} \end{equation*}(3.1988)F3103(h1022)2(f1kHz)2 W/m2
where h h hhh and f f fff are now the amplitude and the frequency, respectively, of the gravitational waves as measured on Earth. Inserting h = 10 21 h = 10 21 h=10^(-21)h=10^{-21}h=1021 and f = 200 Hz f = 200 Hz f=200Hzf=200 \mathrm{~Hz}f=200 Hz for GW150914, we find that
(3.1989) F 3 10 3 10 2 ( 1 5 ) 2 W / m 2 = 12 10 3 W / m 2 10 2 W / m 2 (3.1989) F 3 10 3 10 2 1 5 2 W / m 2 = 12 10 3 W / m 2 10 2 W / m 2 {:(3.1989)F∼3*10^(-3)*10^(2)*((1)/(5))^(2)W//m^(2)=12*10^(-3)W//m^(2)∼10^(-2)W//m^(2):}\begin{equation*} F \sim 3 \cdot 10^{-3} \cdot 10^{2} \cdot\left(\frac{1}{5}\right)^{2} \mathrm{~W} / \mathrm{m}^{2}=12 \cdot 10^{-3} \mathrm{~W} / \mathrm{m}^{2} \sim 10^{-2} \mathrm{~W} / \mathrm{m}^{2} \tag{3.1989} \end{equation*}(3.1989)F3103102(15)2 W/m2=12103 W/m2102 W/m2
which is an enormous amount of energy flux compared to the observed energy flux in electromagnetic waves.
b) To estimate the energy flux in electromagnetic waves that is received at Earth from a full moon, we use the following assumptions: (i) The solar irradiance is about 1387 W / m 2 1387 W / m 2 1387W//m^(2)1387 \mathrm{~W} / \mathrm{m}^{2}1387 W/m2 at the surface of the Moon and the Earth; (ii) The reflectivity of the Moon is about 12 % 12 % 12%12 \%12%; and (iii) The average solid angle subtended by the Moon in the sky at Earth is about 6.4 10 5 6.4 10 5 6.4*10^(-5)6.4 \cdot 10^{-5}6.4105 steradians, which is approximately the same as that subtended by the Sun, i.e., about 6.8 10 5 6.8 10 5 6.8*10^(-5)6.8 \cdot 10^{-5}6.8105 steradians. Therefore, the energy flux from a full moon is given by
(3.1990) F full moon = 1387 W / m 2 0.12 6.4 10 5 = 11 10 3 W / m 2 (3.1990) F full moon  = 1387 W / m 2 0.12 6.4 10 5 = 11 10 3 W / m 2 {:(3.1990)F_("full moon ")=1387W//m^(2)*0.12*6.4*10^(-5)=11*10^(-3)W//m^(2):}\begin{equation*} F_{\text {full moon }}=1387 \mathrm{~W} / \mathrm{m}^{2} \cdot 0.12 \cdot 6.4 \cdot 10^{-5}=11 \cdot 10^{-3} \mathrm{~W} / \mathrm{m}^{2} \tag{3.1990} \end{equation*}(3.1990)Ffull moon =1387 W/m20.126.4105=11103 W/m2
The assumptions made are not totally accurate, so reducing the solar irradiance to 1000 W / m 2 1000 W / m 2 1000W//m^(2)1000 \mathrm{~W} / \mathrm{m}^{2}1000 W/m2 and the reflectivity of the Moon to 10 % 10 % 10%10 \%10% lead to F full moon = 6.4 10 3 F full moon  = 6.4 10 3 F_("full moon ")=6.4*10^(-3)F_{\text {full moon }}=6.4 \cdot 10^{-3}Ffull moon =6.4103 W / m 2 W / m 2 W//m^(2)\mathrm{W} / \mathrm{m}^{2}W/m2. Thus, the gravitational wave energy flux of GW150914 in a), i.e., F GW F GW F_(GW)∼F_{\mathrm{GW}} \simFGW 12 10 3 W / m 2 12 10 3 W / m 2 12*10^(-3)W//m^(2)12 \cdot 10^{-3} \mathrm{~W} / \mathrm{m}^{2}12103 W/m2, is between once or twice the electromagnetic wave energy flux of a full moon [ F GW ( 1 , 2 ) F full moon ] F GW ( 1 , 2 ) F full moon  [F_(GW)∼(1,2)F_("full moon ")]\left[F_{\mathrm{GW}} \sim(1,2) F_{\text {full moon }}\right][FGW(1,2)Ffull moon ], although the estimated distance of GW150914 is about 400 Mpc and the distance from the Earth to the Moon is just 1.25 10 8 pc 1.25 10 8 pc 1.25*10^(-8)pc1.25 \cdot 10^{-8} \mathrm{pc}1.25108pc.

2.137

The metric for a linearly expanding spacetime is given by
(3.1991) d s 2 = d t 2 H 2 t 2 d x 2 (3.1991) d s 2 = d t 2 H 2 t 2 d x 2 {:(3.1991)ds^(2)=dt^(2)-H^(2)t^(2)dx^(2):}\begin{equation*} d s^{2}=d t^{2}-H^{2} t^{2} d x^{2} \tag{3.1991} \end{equation*}(3.1991)ds2=dt2H2t2dx2
We want to start at t = t 0 , x = 0 t = t 0 , x = 0 t=t_(0),x=0t=t_{0}, x=0t=t0,x=0 and arrive at t = t 1 , x = L t = t 1 , x = L t=t_(1),x=Lt=t_{1}, x=Lt=t1,x=L without accelerating. This implies that we should move along a geodesic. Thus, we want to determine x ( t ) x ( t ) x(t)x(t)x(t). The Lagrangian is given by
(3.1992) t ˙ 2 H 2 t 2 x ˙ 2 = β = const. (3.1992) t ˙ 2 H 2 t 2 x ˙ 2 = β =  const.  {:(3.1992)t^(˙)^(2)-H^(2)t^(2)x^(˙)^(2)=beta=" const. ":}\begin{equation*} \dot{t}^{2}-H^{2} t^{2} \dot{x}^{2}=\beta=\text { const. } \tag{3.1992} \end{equation*}(3.1992)t˙2H2t2x˙2=β= const. 
The Euler-Lagrange equations are
(3.1993) d d τ ( H 2 t 2 x ˙ ) = 0 (3.1994) t ¨ H 2 t x ˙ 2 = 0 (3.1993) d d τ H 2 t 2 x ˙ = 0 (3.1994) t ¨ H 2 t x ˙ 2 = 0 {:[(3.1993)(d)/(d tau)(H^(2)t^(2)(x^(˙)))=0],[(3.1994)t^(¨)-H^(2)tx^(˙)^(2)=0]:}\begin{align*} \frac{d}{d \tau}\left(H^{2} t^{2} \dot{x}\right) & =0 \tag{3.1993}\\ \ddot{t}-H^{2} t \dot{x}^{2} & =0 \tag{3.1994} \end{align*}(3.1993)ddτ(H2t2x˙)=0(3.1994)t¨H2tx˙2=0
From the first equation, we obtain
(3.1995) x ˙ = A H 2 t 2 (3.1995) x ˙ = A H 2 t 2 {:(3.1995)x^(˙)=(A)/(H^(2)t^(2)):}\begin{equation*} \dot{x}=\frac{A}{H^{2} t^{2}} \tag{3.1995} \end{equation*}(3.1995)x˙=AH2t2
where A A AAA is a constant. Inserting this in the Lagrangian, we get
(3.1996) t ˙ 2 H 2 t 2 ( A H 2 t 2 ) 2 = t ˙ 2 A 2 H 2 t 2 = β (3.1996) t ˙ 2 H 2 t 2 A H 2 t 2 2 = t ˙ 2 A 2 H 2 t 2 = β {:(3.1996)t^(˙)^(2)-H^(2)t^(2)((A)/(H^(2)t^(2)))^(2)=t^(˙)^(2)-(A^(2))/(H^(2)t^(2))=beta:}\begin{equation*} \dot{t}^{2}-H^{2} t^{2}\left(\frac{A}{H^{2} t^{2}}\right)^{2}=\dot{t}^{2}-\frac{A^{2}}{H^{2} t^{2}}=\beta \tag{3.1996} \end{equation*}(3.1996)t˙2H2t2(AH2t2)2=t˙2A2H2t2=β
which can be solved for
(3.1997) t ˙ = ± β + A 2 H 2 t 2 = ± 1 H t β H 2 t 2 + A 2 (3.1997) t ˙ = ± β + A 2 H 2 t 2 = ± 1 H t β H 2 t 2 + A 2 {:(3.1997)t^(˙)=+-sqrt(beta+(A^(2))/(H^(2)t^(2)))=+-(1)/(Ht)sqrt(betaH^(2)t^(2)+A^(2)):}\begin{equation*} \dot{t}= \pm \sqrt{\beta+\frac{A^{2}}{H^{2} t^{2}}}= \pm \frac{1}{H t} \sqrt{\beta H^{2} t^{2}+A^{2}} \tag{3.1997} \end{equation*}(3.1997)t˙=±β+A2H2t2=±1HtβH2t2+A2
Now, we can determine x ( t ) x ( t ) x(t)x(t)x(t)
(3.1998) d x d t = d x d τ d τ d t = A H 2 t 2 H t β H 2 t 2 + A 2 = A H t β H 2 t 2 + A 2 (3.1998) d x d t = d x d τ d τ d t = A H 2 t 2 H t β H 2 t 2 + A 2 = A H t β H 2 t 2 + A 2 {:(3.1998)(dx)/(dt)=(dx)/(d tau)(d tau)/(dt)=(A)/(H^(2)t^(2))(Ht)/(sqrt(betaH^(2)t^(2)+A^(2)))=(A)/(Htsqrt(betaH^(2)t^(2)+A^(2))):}\begin{equation*} \frac{d x}{d t}=\frac{d x}{d \tau} \frac{d \tau}{d t}=\frac{A}{H^{2} t^{2}} \frac{H t}{\sqrt{\beta H^{2} t^{2}+A^{2}}}=\frac{A}{H t \sqrt{\beta H^{2} t^{2}+A^{2}}} \tag{3.1998} \end{equation*}(3.1998)dxdt=dxdτdτdt=AH2t2HtβH2t2+A2=AHtβH2t2+A2
and integrating on both sides, we obtain
(3.1999) 0 x ( t ) d x = t 0 t 1 A H t β H 2 t 2 + A 2 d t (3.1999) 0 x ( t ) d x = t 0 t 1 A H t β H 2 t 2 + A 2 d t {:(3.1999)int_(0)^(x(t))dx^(')=int_(t_(0))^(t_(1))(A)/(Htsqrt(betaH^(2)t^(2)+A^(2)))dt:}\begin{equation*} \int_{0}^{x(t)} d x^{\prime}=\int_{t_{0}}^{t_{1}} \frac{A}{H t \sqrt{\beta H^{2} t^{2}+A^{2}}} d t \tag{3.1999} \end{equation*}(3.1999)0x(t)dx=t0t1AHtβH2t2+A2dt
and thus, we have
(3.2000) x ( t ) = t 0 t A H t β H 2 t 2 + A 2 d t (3.2000) x ( t ) = t 0 t A H t β H 2 t 2 + A 2 d t {:(3.2000)x(t)=int_(t_(0))^(t)(A)/(Ht^(')sqrt(betaH^(2)t^('2)+A^(2)))dt^('):}\begin{equation*} x(t)=\int_{t_{0}}^{t} \frac{A}{H t^{\prime} \sqrt{\beta H^{2} t^{\prime 2}+A^{2}}} d t^{\prime} \tag{3.2000} \end{equation*}(3.2000)x(t)=t0tAHtβH2t2+A2dt
The farthest one could go, x ( t 1 ) = L x t 1 = L x(t_(1))=Lx\left(t_{1}\right)=Lx(t1)=L would be following a lightlike geodesic, i.e., for β = 0 β = 0 beta=0\beta=0β=0. Then, we obtain
(3.2001) L = t 0 t 1 A H t A 2 d t = t 0 t 1 1 H t d t = 1 H ln ( t 1 t 0 ) (3.2001) L = t 0 t 1 A H t A 2 d t = t 0 t 1 1 H t d t = 1 H ln t 1 t 0 {:(3.2001)L=int_(t_(0))^(t_(1))(A)/(Ht^(')sqrt(A^(2)))dt^(')=int_(t_(0))^(t_(1))(1)/(Ht^('))dt^(')=(1)/(H)ln((t_(1))/(t_(0))):}\begin{equation*} L=\int_{t_{0}}^{t_{1}} \frac{A}{H t^{\prime} \sqrt{A^{2}}} d t^{\prime}=\int_{t_{0}}^{t_{1}} \frac{1}{H t^{\prime}} d t^{\prime}=\frac{1}{H} \ln \left(\frac{t_{1}}{t_{0}}\right) \tag{3.2001} \end{equation*}(3.2001)L=t0t1AHtA2dt=t0t11Htdt=1Hln(t1t0)
2.138
The cosmological redshift is due to the fact that signals interchanged between two observers will have different travel times if the size of the universe is altered. In the case of the 2-dimensional mini-universe, we assume that χ 0 < χ 1 χ 0 < χ 1 chi_(0) < chi_(1)\chi_{0}<\chi_{1}χ0<χ1. (To treat the case of χ 1 > χ 0 χ 1 > χ 0 chi_(1) > chi_(0)\chi_{1}>\chi_{0}χ1>χ0, simply substitute d χ d χ d chid \chidχ with d χ d χ -d chi-d \chidχ and χ 1 χ 0 χ 1 χ 0 chi_(1)-chi_(0)\chi_{1}-\chi_{0}χ1χ0 with χ 0 χ 1 χ 0 χ 1 chi_(0)-chi_(1)\chi_{0}-\chi_{1}χ0χ1.) For a lightlike geodesic, we have d s = 0 d s = 0 ds=0d s=0ds=0, which gives
(3.2002) c d t = S ( t ) d χ t 0 t 1 c d t S ( t ) = χ 0 χ 1 d χ = χ 1 χ 0 (3.2002) c d t = S ( t ) d χ t 0 t 1 c d t S ( t ) = χ 0 χ 1 d χ = χ 1 χ 0 {:(3.2002)cdt=S(t)d chiquad=>quadint_(t_(0))^(t_(1))(cdt)/(S(t))=int_(chi_(0))^(chi_(1))d chi=chi_(1)-chi_(0):}\begin{equation*} c d t=S(t) d \chi \quad \Rightarrow \quad \int_{t_{0}}^{t_{1}} \frac{c d t}{S(t)}=\int_{\chi_{0}}^{\chi_{1}} d \chi=\chi_{1}-\chi_{0} \tag{3.2002} \end{equation*}(3.2002)cdt=S(t)dχt0t1cdtS(t)=χ0χ1dχ=χ1χ0
We now assume that a light signal is sent from χ 0 χ 0 chi_(0)\chi_{0}χ0 at time t t ttt and another at time t + ε t + ε t+epsit+\varepsilont+ε, they are received at χ 1 χ 1 chi_(1)\chi_{1}χ1 at times t t t^(')t^{\prime}t and t + ε t + ε t^(')+epsi^(')t^{\prime}+\varepsilon^{\prime}t+ε. Assuming that ε ε epsi\varepsilonε is small, the redshift is given by
(3.2003) z = λ 1 λ 0 λ 0 = ε ε 1 (3.2003) z = λ 1 λ 0 λ 0 = ε ε 1 {:(3.2003)z=(lambda_(1)-lambda_(0))/(lambda_(0))=(epsi^('))/(epsi)-1:}\begin{equation*} z=\frac{\lambda_{1}-\lambda_{0}}{\lambda_{0}}=\frac{\varepsilon^{\prime}}{\varepsilon}-1 \tag{3.2003} \end{equation*}(3.2003)z=λ1λ0λ0=εε1
For the two different signals, we obtain
(3.2004) t t c d t S ( t ) = χ 1 χ 0 = t + ε t + ε c d t S ( t ) (3.2004) t t c d t S ( t ) = χ 1 χ 0 = t + ε t + ε c d t S ( t ) {:(3.2004)int_(t)^(t^('))(cdt)/(S(t))=chi_(1)-chi_(0)=int_(t+epsi)^(t^(')+epsi^('))(cdt)/(S(t)):}\begin{equation*} \int_{t}^{t^{\prime}} \frac{c d t}{S(t)}=\chi_{1}-\chi_{0}=\int_{t+\varepsilon}^{t^{\prime}+\varepsilon^{\prime}} \frac{c d t}{S(t)} \tag{3.2004} \end{equation*}(3.2004)ttcdtS(t)=χ1χ0=t+εt+εcdtS(t)
Using the fact that ε ε epsi\varepsilonε is small, this can be simplified to
(3.2005) ε S ( t ) = ε S ( t ) ε ε = S ( t ) S ( t ) z = S ( t ) S ( t ) 1 (3.2005) ε S ( t ) = ε S t ε ε = S t S ( t ) z = S t S ( t ) 1 {:(3.2005)(epsi)/(S(t))=(epsi^('))/(S(t^(')))quad=>quad(epsi^('))/(epsi)=(S(t^(')))/(S(t))quad=>quad z=(S(t^(')))/(S(t))-1:}\begin{equation*} \frac{\varepsilon}{S(t)}=\frac{\varepsilon^{\prime}}{S\left(t^{\prime}\right)} \quad \Rightarrow \quad \frac{\varepsilon^{\prime}}{\varepsilon}=\frac{S\left(t^{\prime}\right)}{S(t)} \quad \Rightarrow \quad z=\frac{S\left(t^{\prime}\right)}{S(t)}-1 \tag{3.2005} \end{equation*}(3.2005)εS(t)=εS(t)εε=S(t)S(t)z=S(t)S(t)1

2.139

The metric components are independent of the coordinate x x xxx and thus x x del_(x)\partial_{x}x is a Killing vector field, implying that
(3.2006) k = g ( N , x ) = g x x N x , (3.2006) k = g N , x = g x x N x , {:(3.2006)k=-g(N,del_(x))=-g_(xx)N^(x)",":}\begin{equation*} k=-g\left(N, \partial_{x}\right)=-g_{x x} N^{x}, \tag{3.2006} \end{equation*}(3.2006)k=g(N,x)=gxxNx,
where N N NNN is the 4 -frequency of the light signal, is a constant. Since the 4 -frequency is a null vector, we also obtain
g ( N , N ) = g t t ( N t ) 2 + g x x ( N x ) 2 = 0 N t = g x x ( N x ) 2 g t t = k g t t g x x g ( N , N ) = g t t N t 2 + g x x N x 2 = 0 N t = g x x N x 2 g t t = k g t t g x x g(N,N)=g_(tt)(N^(t))^(2)+g_(xx)(N^(x))^(2)=0quad LongrightarrowquadN^(t)=sqrt(-(g_(xx)(N^(x))^(2))/(g_(tt)))=(k)/(sqrt(-g_(tt)g_(xx)))g(N, N)=g_{t t}\left(N^{t}\right)^{2}+g_{x x}\left(N^{x}\right)^{2}=0 \quad \Longrightarrow \quad N^{t}=\sqrt{-\frac{g_{x x}\left(N^{x}\right)^{2}}{g_{t t}}}=\frac{k}{\sqrt{-g_{t t} g_{x x}}}g(N,N)=gtt(Nt)2+gxx(Nx)2=0Nt=gxx(Nx)2gtt=kgttgxx.
The 4-velocity of a comoving observer is given by V = γ t V = γ t V=gammadel_(t)V=\gamma \partial_{t}V=γt and normalization results in
(3.2008) g ( V , V ) = g t t γ 2 = 1 γ = 1 g t t (3.2008) g ( V , V ) = g t t γ 2 = 1 γ = 1 g t t {:(3.2008)g(V","V)=g_(tt)gamma^(2)=1quad Longrightarrowquad gamma=(1)/(sqrt(g_(tt))):}\begin{equation*} g(V, V)=g_{t t} \gamma^{2}=1 \quad \Longrightarrow \quad \gamma=\frac{1}{\sqrt{g_{t t}}} \tag{3.2008} \end{equation*}(3.2008)g(V,V)=gttγ2=1γ=1gtt
The frequency observed by such a comoving observer is therefore
(3.2009) ω = g ( N , V ) = g t t γ N t = g t t 1 g t t k g t t g x x = k g x x (3.2009) ω = g ( N , V ) = g t t γ N t = g t t 1 g t t k g t t g x x = k g x x {:(3.2009)omega=g(N","V)=g_(tt)gammaN^(t)=g_(tt)(1)/(sqrt(g_(tt)))(k)/(sqrt(-g_(tt)g_(xx)))=(k)/(sqrt(-g_(xx))):}\begin{equation*} \omega=g(N, V)=g_{t t} \gamma N^{t}=g_{t t} \frac{1}{\sqrt{g_{t t}}} \frac{k}{\sqrt{-g_{t t} g_{x x}}}=\frac{k}{\sqrt{-g_{x x}}} \tag{3.2009} \end{equation*}(3.2009)ω=g(N,V)=gttγNt=gtt1gttkgttgxx=kgxx
The ratio between the observed frequency ω 1 ω 1 omega_(1)\omega_{1}ω1 and the emitted frequency ω 0 ω 0 omega_(0)\omega_{0}ω0 is therefore
(3.2010) ω 1 ω 0 = g x x ( t 0 ) g x x ( t 1 ) = e ( t 0 t 1 ) / R (3.2010) ω 1 ω 0 = g x x t 0 g x x t 1 = e t 0 t 1 / R {:(3.2010)(omega_(1))/(omega_(0))=sqrt((g_(xx)(t_(0)))/(g_(xx)(t_(1))))=e^((t_(0)-t_(1))//R):}\begin{equation*} \frac{\omega_{1}}{\omega_{0}}=\sqrt{\frac{g_{x x}\left(t_{0}\right)}{g_{x x}\left(t_{1}\right)}}=e^{\left(t_{0}-t_{1}\right) / R} \tag{3.2010} \end{equation*}(3.2010)ω1ω0=gxx(t0)gxx(t1)=e(t0t1)/R
Since the light signal travels along a null geodesic, its worldline satisfies
(3.2011) d s 2 = d t 2 e 2 t / R d x 2 = 0 d x d t = e t / R (3.2011) d s 2 = d t 2 e 2 t / R d x 2 = 0 d x d t = e t / R {:(3.2011)ds^(2)=dt^(2)-e^(2t//R)dx^(2)=0quad Longrightarrowquad(dx)/(dt)=e^(-t//R):}\begin{equation*} d s^{2}=d t^{2}-e^{2 t / R} d x^{2}=0 \quad \Longrightarrow \quad \frac{d x}{d t}=e^{-t / R} \tag{3.2011} \end{equation*}(3.2011)ds2=dt2e2t/Rdx2=0dxdt=et/R
Integrating this expression with the initial condition x ( t 0 ) = x 0 x t 0 = x 0 x(t_(0))=x_(0)x\left(t_{0}\right)=x_{0}x(t0)=x0 results in
(3.2012) t 1 = R ln ( e t 0 / R x 1 x 0 R ) (3.2012) t 1 = R ln e t 0 / R x 1 x 0 R {:(3.2012)t_(1)=-R ln(e^(-t_(0)//R)-(x_(1)-x_(0))/(R)):}\begin{equation*} t_{1}=-R \ln \left(e^{-t_{0} / R}-\frac{x_{1}-x_{0}}{R}\right) \tag{3.2012} \end{equation*}(3.2012)t1=Rln(et0/Rx1x0R)
and inserting this into the frequency ratio leads to
(3.2013) z = ω 0 ω 1 1 = R e t 0 / R R e t 0 / R x 1 + x 0 1 = x 1 x 0 R e t 0 / R x 1 + x 0 (3.2013) z = ω 0 ω 1 1 = R e t 0 / R R e t 0 / R x 1 + x 0 1 = x 1 x 0 R e t 0 / R x 1 + x 0 {:(3.2013)z=(omega_(0))/(omega_(1))-1=(Re^(-t_(0)//R))/(Re^(-t_(0)//R)-x_(1)+x_(0))-1=(x_(1)-x_(0))/(Re^(-t_(0)//R)-x_(1)+x_(0)):}\begin{equation*} z=\frac{\omega_{0}}{\omega_{1}}-1=\frac{R e^{-t_{0} / R}}{R e^{-t_{0} / R}-x_{1}+x_{0}}-1=\frac{x_{1}-x_{0}}{R e^{-t_{0} / R}-x_{1}+x_{0}} \tag{3.2013} \end{equation*}(3.2013)z=ω0ω11=Ret0/RRet0/Rx1+x01=x1x0Ret0/Rx1+x0

2.140

a) We compute the geodesics for the given metric from the variational principle
(3.2014) δ 1 2 ( t ˙ 2 e 2 t / R x ˙ 2 ) d τ = 0 (3.2014) δ 1 2 t ˙ 2 e 2 t / R x ˙ 2 d τ = 0 {:(3.2014)delta int(1)/(2)(t^(˙)^(2)-e^(2t//R)x^(˙)^(2))d tau=0:}\begin{equation*} \delta \int \frac{1}{2}\left(\dot{t}^{2}-e^{2 t / R} \dot{x}^{2}\right) d \tau=0 \tag{3.2014} \end{equation*}(3.2014)δ12(t˙2e2t/Rx˙2)dτ=0
where we use the same notation as in Problem 2.83. The Euler-Lagrange equations are
(3.2015) t ¨ + 1 R e 2 t / R x ˙ 2 = 0 , e 2 t / R x ¨ 2 R e 2 t / R x ˙ t ˙ = 0 (3.2015) t ¨ + 1 R e 2 t / R x ˙ 2 = 0 , e 2 t / R x ¨ 2 R e 2 t / R x ˙ t ˙ = 0 {:(3.2015)t^(¨)+(1)/(R)*e^(2t//R)x^(˙)^(2)=0","quad-e^(2t//R)x^(¨)-(2)/(R)e^(2t//R)x^(˙)t^(˙)=0:}\begin{equation*} \ddot{t}+\frac{1}{R} \cdot e^{2 t / R} \dot{x}^{2}=0, \quad-e^{2 t / R} \ddot{x}-\frac{2}{R} e^{2 t / R} \dot{x} \dot{t}=0 \tag{3.2015} \end{equation*}(3.2015)t¨+1Re2t/Rx˙2=0,e2t/Rx¨2Re2t/Rx˙t˙=0
and by comparing this with the general form of the geodesics equation
(3.2016) x ¨ μ + Γ α β μ x ˙ α x ˙ β = 0 , (3.2016) x ¨ μ + Γ α β μ x ˙ α x ˙ β = 0 , {:(3.2016)x^(¨)^(mu)+Gamma_(alpha beta)^(mu)x^(˙)^(alpha)x^(˙)^(beta)=0",":}\begin{equation*} \ddot{x}^{\mu}+\Gamma_{\alpha \beta}^{\mu} \dot{x}^{\alpha} \dot{x}^{\beta}=0, \tag{3.2016} \end{equation*}(3.2016)x¨μ+Γαβμx˙αx˙β=0,
we can identify the nonzero Christoffel symbols
(3.2017) Γ x x t = 1 R e 2 t / R , Γ t x x = Γ x t x = 1 R (3.2017) Γ x x t = 1 R e 2 t / R , Γ t x x = Γ x t x = 1 R {:(3.2017)Gamma_(xx)^(t)=(1)/(R)e^(2t//R)","quadGamma_(tx)^(x)=Gamma_(xt)^(x)=(1)/(R):}\begin{equation*} \Gamma_{x x}^{t}=\frac{1}{R} e^{2 t / R}, \quad \Gamma_{t x}^{x}=\Gamma_{x t}^{x}=\frac{1}{R} \tag{3.2017} \end{equation*}(3.2017)Γxxt=1Re2t/R,Γtxx=Γxtx=1R
b) The metric tensor is
(3.2018) ( g μ ν ) = ( 1 0 0 e 2 t / R ) (3.2018) g μ ν = 1 0 0 e 2 t / R {:(3.2018)(g_(mu nu))=([1,0],[0,-e^(2t//R)]):}\left(g_{\mu \nu}\right)=\left(\begin{array}{cc} 1 & 0 \tag{3.2018}\\ 0 & -e^{2 t / R} \end{array}\right)(3.2018)(gμν)=(100e2t/R)
and thus, we have the inverse metric tensor
(3.2019) ( g μ v ) = ( 1 0 0 e 2 t / R ) (3.2019) g μ v = 1 0 0 e 2 t / R {:(3.2019)(g^(mu v))=([1,0],[0,-e^(-2t//R)]):}\left(g^{\mu v}\right)=\left(\begin{array}{cc} 1 & 0 \tag{3.2019}\\ 0 & -e^{-2 t / R} \end{array}\right)(3.2019)(gμv)=(100e2t/R)
We find that
g μ ν μ v Φ = ν v Φ = ν ν Φ + Γ v α v α Φ (3.2020) = g μ ν μ ν Φ + Γ ν α v g α β β Φ = Φ t t e 2 t / R Φ x x + 1 R Φ t = Γ x t g t h Φ t = 0 , g μ ν μ v Φ = ν v Φ = ν ν Φ + Γ v α v α Φ (3.2020) = g μ ν μ ν Φ + Γ ν α v g α β β Φ = Φ t t e 2 t / R Φ x x + 1 R Φ t = Γ x t g t h Φ t = 0 , {:[g_(mu nu)grad^(mu)grad^(v)Phi=grad^(nu)del_(v)Phi=del^(nu)del_(nu)Phi+Gamma_(v alpha)^(v)del^(alpha)Phi],[(3.2020)=g^(mu nu)del_(mu)del_(nu)Phi+Gamma_(nu alpha)^(v)g^(alpha beta)del_(beta)Phi=Phi_(tt)-e^(-2t//R)Phi_(xx)+ubrace((1)/(R)Phi_(t)ubrace)_(=Gamma_(xt)^(**)g^(th)Phi_(t))=0","]:}\begin{align*} g_{\mu \nu} \nabla^{\mu} \nabla^{v} \Phi & =\nabla^{\nu} \partial_{v} \Phi=\partial^{\nu} \partial_{\nu} \Phi+\Gamma_{v \alpha}^{v} \partial^{\alpha} \Phi \\ & =g^{\mu \nu} \partial_{\mu} \partial_{\nu} \Phi+\Gamma_{\nu \alpha}^{v} g^{\alpha \beta} \partial_{\beta} \Phi=\Phi_{t t}-e^{-2 t / R} \Phi_{x x}+\underbrace{\frac{1}{R} \Phi_{t}}_{=\Gamma_{x t}^{*} g^{t h} \Phi_{t}}=0, \tag{3.2020} \end{align*}gμνμvΦ=νvΦ=ννΦ+ΓvαvαΦ(3.2020)=gμνμνΦ+ΓναvgαββΦ=Φtte2t/RΦxx+1RΦt=ΓxtgthΦt=0,
where Φ t = Φ / t Φ t = Φ / t Phi_(t)=del Phi//del t\Phi_{t}=\partial \Phi / \partial tΦt=Φ/t, etc.

2.141

a) The geodesic equation can be computed from t ˙ 2 exp ( 2 t / t H ) r ˙ 2 = 0 t ˙ 2 exp 2 t / t H r ˙ 2 = 0 t^(˙)^(2)-exp(2t//t_(H))r^(˙)^(2)=0\dot{t}^{2}-\exp \left(2 t / t_{H}\right) \dot{r}^{2}=0t˙2exp(2t/tH)r˙2=0, where t ˙ = d t / d τ t ˙ = d t / d τ t^(˙)=dt//d tau\dot{t}=d t / d \taut˙=dt/dτ, i.e., d r / d t = ± exp ( t / t H ) d r / d t = ± exp t / t H dr//dt=+-exp(-t//t_(H))d r / d t= \pm \exp \left(-t / t_{H}\right)dr/dt=±exp(t/tH) with r ( t 0 ) = 0 r t 0 = 0 r(t_(0))=0r\left(t_{0}\right)=0r(t0)=0. Using d r / d t = d r / d t = dr//dt=d r / d t=dr/dt= + exp ( t / t H ) > 0 + exp t / t H > 0 +exp(-t//t_(H)) > 0+\exp \left(-t / t_{H}\right)>0+exp(t/tH)>0, this implies that
(3.2021) r ( t ) = 0 r d r = t 0 t e t / t H d t = [ t H e t / t H ] t 0 t = t H ( e t / t H e t 0 / t H ) (3.2021) r ( t ) = 0 r d r = t 0 t e t / t H d t = t H e t / t H t 0 t = t H e t / t H e t 0 / t H {:(3.2021)r(t)=int_(0)^(r)dr^(')=int_(t_(0))^(t)e^(-t^(')//t_(H))dt^(')=[-t_(H)e^(-t^(')//t_(H))]_(t_(0))^(t)=-t_(H)(e^(-t//t_(H))-e^(-t_(0)//t_(H))):}\begin{equation*} r(t)=\int_{0}^{r} d r^{\prime}=\int_{t_{0}}^{t} e^{-t^{\prime} / t_{H}} d t^{\prime}=\left[-t_{H} e^{-t^{\prime} / t_{H}}\right]_{t_{0}}^{t}=-t_{H}\left(e^{-t / t_{H}}-e^{-t_{0} / t_{H}}\right) \tag{3.2021} \end{equation*}(3.2021)r(t)=0rdr=t0tet/tHdt=[tHet/tH]t0t=tH(et/tHet0/tH)
b) For the line θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2 and φ = 0 φ = 0 varphi=0\varphi=0φ=0 at fixed universal time t = t 0 = t = t 0 = t=t_(0)=t=t_{0}=t=t0= const., we have d t = 0 d t = 0 dt=0d t=0dt=0 and d Ω = 0 d Ω = 0 d Omega=0d \Omega=0dΩ=0, which imply that
(3.2022) d s 2 = e 2 t / t H d r 2 g r r ( t ) = e 2 t / t H (3.2022) d s 2 = e 2 t / t H d r 2 g r r ( t ) = e 2 t / t H {:(3.2022)ds^(2)=-e^(2t//t_(H))dr^(2)quad=>quadg_(rr)(t)=-e^(2t//t_(H)):}\begin{equation*} d s^{2}=-e^{2 t / t_{H}} d r^{2} \quad \Rightarrow \quad g_{r r}(t)=-e^{2 t / t_{H}} \tag{3.2022} \end{equation*}(3.2022)ds2=e2t/tHdr2grr(t)=e2t/tH
Thus, the proper distance between the origin r = 0 r = 0 r=0r=0r=0 and a point r > 0 r > 0 r > 0r>0r>0 is given by
(3.2023) d p ( t , r ) = r = 0 r = r g r r ( t ) d r = 0 r e 2 t / t H d r = e t / t H 0 r d r = r e t / t H (3.2023) d p ( t , r ) = r = 0 r = r g r r ( t ) d r = 0 r e 2 t / t H d r = e t / t H 0 r d r = r e t / t H {:(3.2023)d_(p)(t","r)=int_(r^(')=0)^(r^(')=r)sqrt(-g_(rr)(t))dr^(')=int_(0)^(r)sqrt(e^(2t//t_(H)))dr^(')=e^(t//t_(H))int_(0)^(r)dr^(')=re^(t//t_(H)):}\begin{equation*} d_{p}(t, r)=\int_{r^{\prime}=0}^{r^{\prime}=r} \sqrt{-g_{r r}(t)} d r^{\prime}=\int_{0}^{r} \sqrt{e^{2 t / t_{H}}} d r^{\prime}=e^{t / t_{H}} \int_{0}^{r} d r^{\prime}=r e^{t / t_{H}} \tag{3.2023} \end{equation*}(3.2023)dp(t,r)=r=0r=rgrr(t)dr=0re2t/tHdr=et/tH0rdr=ret/tH
c) The cosmological redshift relates the emitted and received wavelengths as (see solutions to Problems 2.138 and 2.139)
(3.2024) λ rec λ em = S ( t rec ) S ( t em ) = e t rec / t H e t em / t H = e ( t rec t em ) / t H (3.2024) λ rec λ em = S t rec S t em = e t rec / t H e t em / t H = e t rec t em / t H {:(3.2024)(lambda_(rec))/(lambda_(em))=(S(t_(rec)))/(S(t_(em)))=(e^(t_(rec)//t_(H)))/(e^(t_(em)//t_(H)))=e^((t_(rec)-t_(em))//t_(H)):}\begin{equation*} \frac{\lambda_{\mathrm{rec}}}{\lambda_{\mathrm{em}}}=\frac{S\left(t_{\mathrm{rec}}\right)}{S\left(t_{\mathrm{em}}\right)}=\frac{e^{t_{\mathrm{rec}} / t_{H}}}{e^{t_{\mathrm{em}} / t_{H}}}=e^{\left(t_{\mathrm{rec}}-t_{\mathrm{em}}\right) / t_{H}} \tag{3.2024} \end{equation*}(3.2024)λrecλem=S(trec)S(tem)=etrec/tHetem/tH=e(trectem)/tH
Thus, using t em = t 0 t em = t 0 t_(em)=t_(0)t_{\mathrm{em}}=t_{0}tem=t0 and t rec = t t rec = t t_(rec)=tt_{\mathrm{rec}}=ttrec=t, the spectral shift of the light ray is given by
(3.2025) z λ rec λ em 1 = e ( t t 0 ) / t H 1 (3.2025) z λ rec λ em 1 = e t t 0 / t H 1 {:(3.2025)z-=(lambda_(rec))/(lambda_(em))-1=e^((t-t_(0))//t_(H))-1:}\begin{equation*} z \equiv \frac{\lambda_{\mathrm{rec}}}{\lambda_{\mathrm{em}}}-1=e^{\left(t-t_{0}\right) / t_{H}}-1 \tag{3.2025} \end{equation*}(3.2025)zλrecλem1=e(tt0)/tH1

2.142

a) At cosmological time t t ttt we find that
(3.2026) g t t = 1 , g i j = a ( t ) 2 G i j (3.2026) g t t = 1 , g i j = a ( t ) 2 G i j {:(3.2026)g_(tt)=1","quadg_(ij)=-a(t)^(2)G_(ij):}\begin{equation*} g_{t t}=1, \quad g_{i j}=-a(t)^{2} G_{i j} \tag{3.2026} \end{equation*}(3.2026)gtt=1,gij=a(t)2Gij
where G i j G i j G_(ij)G_{i j}Gij are the components of a Riemannian metric. The proper length of a curve on the cosmological simultaneity is given by
(3.2027) = g i j x ˙ i x ˙ j d s = a ( t ) G i j x ˙ i x ˙ j d s a ( t ) c , (3.2027) = g i j x ˙ i x ˙ j d s = a ( t ) G i j x ˙ i x ˙ j d s a ( t ) c , {:(3.2027)ℓ=intsqrt(-g_(ij)x^(˙)^(i)x^(˙)^(j))ds=a(t)intsqrt(G_(ij)x^(˙)^(i)x^(˙)^(j))ds-=a(t)ℓ_(c)",":}\begin{equation*} \ell=\int \sqrt{-g_{i j} \dot{x}^{i} \dot{x}^{j}} d s=a(t) \int \sqrt{G_{i j} \dot{x}^{i} \dot{x}^{j}} d s \equiv a(t) \ell_{c}, \tag{3.2027} \end{equation*}(3.2027)=gijx˙ix˙jds=a(t)Gijx˙ix˙jdsa(t)c,
where s s sss is the curve parameter and c c ℓ_(c)\ell_{c}c the comoving length of the curve. Fixing the end points of the curve and minimizing this length gives the proper distance d p d p d_(p)d_{p}dp between the points as d p = a ( t ) d c d p = a ( t ) d c d_(p)=a(t)d_(c)d_{p}=a(t) d_{c}dp=a(t)dc, where the comoving distance d c d c d_(c)d_{c}dc is the minimal value of the integral
(3.2028) d c = min ( G i j x ˙ i x ˙ j d s ) (3.2028) d c = min G i j x ˙ i x ˙ j d s {:(3.2028)d_(c)=min(intsqrt(G_(ij)x^(˙)^(i)x^(˙)^(j))ds):}\begin{equation*} d_{c}=\min \left(\int \sqrt{G_{i j} \dot{x}^{i} \dot{x}^{j}} d s\right) \tag{3.2028} \end{equation*}(3.2028)dc=min(Gijx˙ix˙jds)
which is independent of the cosmological time t t ttt. It follows that
(3.2029) v p = d ( d p ) d t = a ˙ ( t ) d c = a ˙ a d p (3.2029) v p = d d p d t = a ˙ ( t ) d c = a ˙ a d p {:(3.2029)v_(p)=(d(d_(p)))/(dt)=a^(˙)(t)d_(c)=((a^(˙)))/(a)d_(p):}\begin{equation*} v_{p}=\frac{d\left(d_{p}\right)}{d t}=\dot{a}(t) d_{c}=\frac{\dot{a}}{a} d_{p} \tag{3.2029} \end{equation*}(3.2029)vp=d(dp)dt=a˙(t)dc=a˙adp
b) Hubble's law describes the observation that objects (read: galaxies) are moving away from the Earth at velocities proportional to their distances. The further the galaxies are, the faster they are moving away. The velocities of the galaxies are measured by their redshifts. The velocities arise from the expansion of the universe. Indeed, the proper distance between galaxies increases even though they are comoving.

2.143

a) If we use the given approximation, i.e., k k kkk is small and ρ ρ Λ ρ ρ Λ rho≃rho_(Lambda)\rho \simeq \rho_{\Lambda}ρρΛ, we can approximate the first Friedmann equation as
(3.2030) a ˙ 2 a 2 = 8 π G 3 ρ Λ (3.2030) a ˙ 2 a 2 = 8 π G 3 ρ Λ {:(3.2030)(a^(˙)^(2))/(a^(2))=(8pi G)/(3)rho_(Lambda):}\begin{equation*} \frac{\dot{a}^{2}}{a^{2}}=\frac{8 \pi G}{3} \rho_{\Lambda} \tag{3.2030} \end{equation*}(3.2030)a˙2a2=8πG3ρΛ
This equation has the solution
(3.2031) a ( t ) = a 0 e ( t t 0 ) / Δ τ , Δ τ = 3 8 π G ρ Λ (3.2031) a ( t ) = a 0 e t t 0 / Δ τ , Δ τ = 3 8 π G ρ Λ {:(3.2031)a(t)=a_(0)e^((t-t_(0))//Delta tau)","quad Delta tau=sqrt((3)/(8pi Grho_(Lambda))):}\begin{equation*} a(t)=a_{0} e^{\left(t-t_{0}\right) / \Delta \tau}, \quad \Delta \tau=\sqrt{\frac{3}{8 \pi G \rho_{\Lambda}}} \tag{3.2031} \end{equation*}(3.2031)a(t)=a0e(tt0)/Δτ,Δτ=38πGρΛ
Inserting this solution into the expression for Ω Ω Omega\OmegaΩ, we obtain
(3.2032) 1 Ω e 2 ( t t 0 ) / Δ τ (3.2032) 1 Ω e 2 t t 0 / Δ τ {:(3.2032)1-Omega prope^(-2(t-t_(0))//Delta tau):}\begin{equation*} 1-\Omega \propto e^{-2\left(t-t_{0}\right) / \Delta \tau} \tag{3.2032} \end{equation*}(3.2032)1Ωe2(tt0)/Δτ
We observe that the larger t t ttt becomes, the smaller 1 Ω 1 Ω 1-Omega1-\Omega1Ω becomes, and thus, 1 Ω 1 Ω 1-Omega1-\Omega1Ω gets closer and closer to zero, which means that Ω Ω Omega\OmegaΩ gets closer and closer to one.
b) Today, we have a universe with a curvature that is very close to zero, which without inflation would need a lot of fine-tuning to maintain. After inflation, a spacetime is obtained, which is very close to being flat as per the solution to a), even if the initial spacetime had a lot of curvature.

2.144

a) The scaling of the energy density for an ideal fluid with equation-of-state parameter w w www is given by ρ = ρ 0 a 3 ( 1 + w ) ρ = ρ 0 a 3 ( 1 + w ) rho=rho_(0)a^(-3(1+w))\rho=\rho_{0} a^{-3(1+w)}ρ=ρ0a3(1+w). The density of component i i iii today is given by
(3.2033) ρ 0 , i = ρ c Ω i (3.2033) ρ 0 , i = ρ c Ω i {:(3.2033)rho_(0,i)=rho_(c)Omega_(i):}\begin{equation*} \rho_{0, i}=\rho_{c} \Omega_{i} \tag{3.2033} \end{equation*}(3.2033)ρ0,i=ρcΩi
where ρ c ρ c rho_(c)\rho_{c}ρc is the critical density today. This means that
ρ 0 , Λ = ρ c Ω Λ 0.7 ρ c ρ Λ = 0.7 ρ c a 3 ( 1 1 ) = 0.7 ρ c , ρ 0 , m = ρ c Ω m 0.3 ρ c ρ m = 0.3 ρ c a 3 , ρ 0 , Λ = ρ c Ω Λ 0.7 ρ c ρ Λ = 0.7 ρ c a 3 ( 1 1 ) = 0.7 ρ c , ρ 0 , m = ρ c Ω m 0.3 ρ c ρ m = 0.3 ρ c a 3 , {:[rho_(0,Lambda)=rho_(c)Omega_(Lambda)≃0.7rho_(c)quad rarrquadrho_(Lambda)=0.7rho_(c)a^(-3(1-1))=0.7rho_(c)","],[rho_(0,m)=rho_(c)Omega_(m)≃0.3rho_(c)quad rarrquadrho_(m)=0.3rho_(c)a^(-3)","]:}\begin{aligned} & \rho_{0, \Lambda}=\rho_{c} \Omega_{\Lambda} \simeq 0.7 \rho_{c} \quad \rightarrow \quad \rho_{\Lambda}=0.7 \rho_{c} a^{-3(1-1)}=0.7 \rho_{c}, \\ & \rho_{0, m}=\rho_{c} \Omega_{m} \simeq 0.3 \rho_{c} \quad \rightarrow \quad \rho_{m}=0.3 \rho_{c} a^{-3}, \end{aligned}ρ0,Λ=ρcΩΛ0.7ρcρΛ=0.7ρca3(11)=0.7ρc,ρ0,m=ρcΩm0.3ρcρm=0.3ρca3,
(3.2034) ρ 0 , r = ρ c Ω r 10 4 ρ c ρ r = 10 4 ρ c a 3 ( 1 + 1 3 ) = 10 4 ρ c a 4 , (3.2034) ρ 0 , r = ρ c Ω r 10 4 ρ c ρ r = 10 4 ρ c a 3 1 + 1 3 = 10 4 ρ c a 4 , {:(3.2034)rho_(0,r)=rho_(c)Omega_(r)≃10^(-4)rho_(c)quad rarrquadrho_(r)=10^(-4)rho_(c)a^(-3(1+(1)/(3)))=10^(-4)rho_(c)a^(-4)",":}\begin{equation*} \rho_{0, r}=\rho_{c} \Omega_{r} \simeq 10^{-4} \rho_{c} \quad \rightarrow \quad \rho_{r}=10^{-4} \rho_{c} a^{-3\left(1+\frac{1}{3}\right)}=10^{-4} \rho_{c} a^{-4}, \tag{3.2034} \end{equation*}(3.2034)ρ0,r=ρcΩr104ρcρr=104ρca3(1+13)=104ρca4,
since w = 1 w = 1 w=-1w=-1w=1 for a cosmological constant, w = 0 w = 0 w=0w=0w=0 for a matter gas, and w = 1 3 w = 1 3 w=(1)/(3)w=\frac{1}{3}w=13 for a radiation gas. The ratio between the energy densities of matter and the cosmological constant is therefore given by
(3.2035) ρ Λ ρ m = Ω Λ Ω m 1 a 3 . (3.2035) ρ Λ ρ m = Ω Λ Ω m 1 a 3 . {:(3.2035)(rho_(Lambda))/(rho_(m))=(Omega_(Lambda))/(Omega_(m))(1)/(a^(-3)).:}\begin{equation*} \frac{\rho_{\Lambda}}{\rho_{m}}=\frac{\Omega_{\Lambda}}{\Omega_{m}} \frac{1}{a^{-3}} . \tag{3.2035} \end{equation*}(3.2035)ρΛρm=ΩΛΩm1a3.
The energy densities were therefore the same when this ratio was equal to one, i.e., when
(3.2036) a 3 = Ω m Ω Λ a = ( Ω m Ω Λ ) 1 / 3 ( 0.3 0.7 ) 1 / 3 0.75 . (3.2036) a 3 = Ω m Ω Λ a = Ω m Ω Λ 1 / 3 0.3 0.7 1 / 3 0.75 . {:(3.2036)a^(3)=(Omega_(m))/(Omega_(Lambda))=>a=((Omega_(m))/(Omega_(Lambda)))^(1//3)≃((0.3)/(0.7))^(1//3)≃0.75.:}\begin{equation*} a^{3}=\frac{\Omega_{m}}{\Omega_{\Lambda}} \Rightarrow a=\left(\frac{\Omega_{m}}{\Omega_{\Lambda}}\right)^{1 / 3} \simeq\left(\frac{0.3}{0.7}\right)^{1 / 3} \simeq 0.75 . \tag{3.2036} \end{equation*}(3.2036)a3=ΩmΩΛa=(ΩmΩΛ)1/3(0.30.7)1/30.75.
Note that this is not very long ago. A naïve estimate with H 0 70 km / s / Mpc H 0 70 km / s / Mpc H_(0)≃70km//s//Mpc≃H_{0} \simeq 70 \mathrm{~km} / \mathrm{s} / \mathrm{Mpc} \simeqH070 km/s/Mpc 2 10 18 s 1 2 10 18 s 1 2*10^(-18)s^(-1)2 \cdot 10^{-18} \mathrm{~s}^{-1}21018 s1 gives a couple of billion years ago (similar to the birth of the solar system).
b) Similarly, the ratio between matter and radiation energy densities is given by
(3.2037) ρ m ρ r = Ω m Ω r a 3 a 4 = Ω m Ω r a . (3.2037) ρ m ρ r = Ω m Ω r a 3 a 4 = Ω m Ω r a . {:(3.2037)(rho_(m))/(rho_(r))=(Omega_(m))/(Omega_(r))(a^(-3))/(a^(-4))=(Omega_(m))/(Omega_(r))a.:}\begin{equation*} \frac{\rho_{m}}{\rho_{r}}=\frac{\Omega_{m}}{\Omega_{r}} \frac{a^{-3}}{a^{-4}}=\frac{\Omega_{m}}{\Omega_{r}} a . \tag{3.2037} \end{equation*}(3.2037)ρmρr=ΩmΩra3a4=ΩmΩra.
This ratio is equal to one when
(3.2038) a = Ω r Ω m 10 4 0.3 = 1 3000 . (3.2038) a = Ω r Ω m 10 4 0.3 = 1 3000 . {:(3.2038)a=(Omega_(r))/(Omega_(m))≃(10^(-4))/(0.3)=(1)/(3000).:}\begin{equation*} a=\frac{\Omega_{r}}{\Omega_{m}} \simeq \frac{10^{-4}}{0.3}=\frac{1}{3000} . \tag{3.2038} \end{equation*}(3.2038)a=ΩrΩm1040.3=13000.
As the cosmological redshift is given by 1 + z = a 0 a 1 + z = a 0 a 1+z=(a_(0))/(a)1+z=\frac{a_{0}}{a}1+z=a0a, we find that
(3.2039) z = 1 a 1 3000 , (3.2039) z = 1 a 1 3000 , {:(3.2039)z=(1)/(a)-1≃3000",":}\begin{equation*} z=\frac{1}{a}-1 \simeq 3000, \tag{3.2039} \end{equation*}(3.2039)z=1a13000,
at matter-radiation equality.

2.145

From the relation
(3.2040) a ˙ = 8 π G 3 i ρ 0 i a 1 3 w i , (3.2040) a ˙ = 8 π G 3 i ρ 0 i a 1 3 w i , {:(3.2040)a^(˙)=sqrt((8pi G)/(3))sqrt(sum_(i)rho_(0i)a^(-1-3w_(i)))",":}\begin{equation*} \dot{a}=\sqrt{\frac{8 \pi G}{3}} \sqrt{\sum_{i} \rho_{0 i} a^{-1-3 w_{i}}}, \tag{3.2040} \end{equation*}(3.2040)a˙=8πG3iρ0ia13wi,
with the definitions
(3.2041) ρ c = 3 H 2 8 π G and ρ 0 i = ρ 0 c Ω i , (3.2041) ρ c = 3 H 2 8 π G  and  ρ 0 i = ρ 0 c Ω i , {:(3.2041)rho_(c)=(3H^(2))/(8pi G)quad" and "quadrho_(0i)=rho_(0c)Omega_(i)",":}\begin{equation*} \rho_{c}=\frac{3 H^{2}}{8 \pi G} \quad \text { and } \quad \rho_{0 i}=\rho_{0 c} \Omega_{i}, \tag{3.2041} \end{equation*}(3.2041)ρc=3H28πG and ρ0i=ρ0cΩi,
Figure 3.18 The scale factor a ( t ) a ( t ) a(t)a(t)a(t) for the solution to Problem 2.145. The solution for Ω Λ = 1 Ω Λ = 1 Omega_(Lambda)=1\Omega_{\Lambda}=1ΩΛ=1 and Ω m = 0 Ω m = 0 Omega_(m)=0\Omega_{m}=0Ωm=0 is also shown for times t > t 0 t > t 0 t > t_(0)t>t_{0}t>t0 for comparison.
we find that
(3.2042) H 0 2 = 8 π G ρ 0 c 3 a ˙ = H 0 i Ω i a 1 3 w i (3.2042) H 0 2 = 8 π G ρ 0 c 3 a ˙ = H 0 i Ω i a 1 3 w i {:(3.2042)H_(0)^(2)=(8pi Grho_(0c))/(3)Longrightarrowa^(˙)=H_(0)sqrt(sum_(i)Omega_(i)a^(-1-3w_(i))):}\begin{equation*} H_{0}^{2}=\frac{8 \pi G \rho_{0 c}}{3} \Longrightarrow \dot{a}=H_{0} \sqrt{\sum_{i} \Omega_{i} a^{-1-3 w_{i}}} \tag{3.2042} \end{equation*}(3.2042)H02=8πGρ0c3a˙=H0iΩia13wi
In our given scenario, there are two components, i = m = i = m = i=m=i=m=i=m= matter and i = Λ = i = Λ = i=Lambda=i=\Lambda=i=Λ= dark energy with corresponding parameters
(3.2043) Ω m = 0.3 , Ω Λ = 0.7 , w m = 0 , w Λ = 1 (3.2043) Ω m = 0.3 , Ω Λ = 0.7 , w m = 0 , w Λ = 1 {:(3.2043)Omega_(m)=0.3","quadOmega_(Lambda)=0.7","quadw_(m)=0","quadw_(Lambda)=-1:}\begin{equation*} \Omega_{m}=0.3, \quad \Omega_{\Lambda}=0.7, \quad w_{m}=0, \quad w_{\Lambda}=-1 \tag{3.2043} \end{equation*}(3.2043)Ωm=0.3,ΩΛ=0.7,wm=0,wΛ=1
The differential equation governing the scale factor is therefore
(3.2044) H 0 = a ˙ 0.3 a 1 + 0.7 a 2 (3.2044) H 0 = a ˙ 0.3 a 1 + 0.7 a 2 {:(3.2044)H_(0)=((a^(˙)))/(sqrt(0.3a^(-1)+0.7a^(2))):}\begin{equation*} H_{0}=\frac{\dot{a}}{\sqrt{0.3 a^{-1}+0.7 a^{2}}} \tag{3.2044} \end{equation*}(3.2044)H0=a˙0.3a1+0.7a2
Integrating this from t 0 t 0 t_(0)t_{0}t0 to t t ttt, we find that
(3.2045) H 0 ( t t 0 ) = 1 a ( t ) d a 0.3 a 1 + 0.7 a 2 (3.2045) H 0 t t 0 = 1 a ( t ) d a 0.3 a 1 + 0.7 a 2 {:(3.2045)H_(0)(t-t_(0))=int_(1)^(a(t))(da)/(sqrt(0.3a^(-1)+0.7a^(2))):}\begin{equation*} H_{0}\left(t-t_{0}\right)=\int_{1}^{a(t)} \frac{d a}{\sqrt{0.3 a^{-1}+0.7 a^{2}}} \tag{3.2045} \end{equation*}(3.2045)H0(tt0)=1a(t)da0.3a1+0.7a2
The result of numerically evaluating this expression is shown in Figure 3.18. Note that the solution quickly becomes dominated by the cosmological constant after t = t 0 t = t 0 t=t_(0)t=t_{0}t=t0 as the matter component dilutes. However, as the cosmological constant is smaller by a factor of 0.7 compared to the case where Ω Λ = 1 Ω Λ = 1 Omega_(Lambda)=1\Omega_{\Lambda}=1ΩΛ=1, the exponential factor in the large time limit (and therefore, H ( t ) H ( t ) H(t rarr oo)H(t \rightarrow \infty)H(t) ) is also smaller by a factor 0.7 .
Unlike the Ω Λ = 1 Ω Λ = 1 Omega_(Lambda)=1\Omega_{\Lambda}=1ΩΛ=1 solution, our solution has a time t t ttt for which a ( t ) = 0 a ( t ) = 0 a(t)=0a(t)=0a(t)=0. Defining this time as t = 0 t = 0 t=0t=0t=0, we find that (from looking at where the graph intersects zero)
(3.2046) H 0 t 0 0.96 t 0 0.96 H 0 (3.2046) H 0 t 0 0.96 t 0 0.96 H 0 {:(3.2046)-H_(0)t_(0)≃-0.96quad Longrightarrowquadt_(0)≃(0.96)/(H_(0)):}\begin{equation*} -H_{0} t_{0} \simeq-0.96 \quad \Longrightarrow \quad t_{0} \simeq \frac{0.96}{H_{0}} \tag{3.2046} \end{equation*}(3.2046)H0t00.96t00.96H0
Current experimental data restricts H 0 H 0 H_(0)H_{0}H0 to be around 70 km / s / Mpc 70 km / s / Mpc 70km//s//Mpc70 \mathrm{~km} / \mathrm{s} / \mathrm{Mpc}70 km/s/Mpc or, equivalently, H 0 7 10 11 yr 1 H 0 7 10 11 yr 1 H_(0)≃7*10^(-11)yr^(-1)H_{0} \simeq 7 \cdot 10^{-11} \mathrm{yr}^{-1}H071011yr1, leading to
(3.2047) t 0 1.3 10 10 years (3.2047) t 0 1.3 10 10  years  {:(3.2047)t_(0)≃1.3*10^(10)" years ":}\begin{equation*} t_{0} \simeq 1.3 \cdot 10^{10} \text { years } \tag{3.2047} \end{equation*}(3.2047)t01.31010 years 
estimating the universe to be around 13 billion years old.

2.146

a) The universe has a Killing vector field φ φ del_(varphi)\partial_{\varphi}φ. Therefore, the quantity g ( φ , γ ˙ ) = g φ , γ ˙ = g(del_(varphi),(gamma^(˙)))=g\left(\partial_{\varphi}, \dot{\gamma}\right)=g(φ,γ˙)= a 2 φ ˙ = a 2 φ ˙ = -a^(2)varphi^(˙)=-a^{2} \dot{\varphi}=a2φ˙= constant = k = k =-k=-k=k. The 4 -velocity of a comoving observer is given by U = t U = t U=del_(t)U=\partial_{t}U=t, and therefore, at t = t 0 t = t 0 t=t_(0)t=t_{0}t=t0
(3.2048) g ( U , γ ˙ ) = 1 1 v 2 = t ˙ (3.2048) g ( U , γ ˙ ) = 1 1 v 2 = t ˙ {:(3.2048)g(U","gamma^(˙))=(1)/(sqrt(1-v^(2)))=t^(˙):}\begin{equation*} g(U, \dot{\gamma})=\frac{1}{\sqrt{1-v^{2}}}=\dot{t} \tag{3.2048} \end{equation*}(3.2048)g(U,γ˙)=11v2=t˙
At t = t 0 t = t 0 t=t_(0)t=t_{0}t=t0, we also have
g ( γ ˙ , γ ˙ ) = 1 = t ˙ 2 a 0 2 φ ˙ 2 = 1 1 v 2 a 0 2 φ ˙ 2 = 1 1 v 2 k a 0 2 (3.2049) k 2 a 0 2 = 1 1 v 2 1 = v 2 1 v 2 k = v a 0 1 v 2 g ( γ ˙ , γ ˙ ) = 1 = t ˙ 2 a 0 2 φ ˙ 2 = 1 1 v 2 a 0 2 φ ˙ 2 = 1 1 v 2 k a 0 2 (3.2049) k 2 a 0 2 = 1 1 v 2 1 = v 2 1 v 2 k = v a 0 1 v 2 {:[g(gamma^(˙)","gamma^(˙))=1=t^(˙)^(2)-a_(0)^(2)varphi^(˙)^(2)=(1)/(1-v^(2))-a_(0)^(2)varphi^(˙)^(2)=(1)/(1-v^(2))-(k)/(a_(0)^(2))],[(3.2049)=>quad(k^(2))/(a_(0)^(2))=(1)/(1-v^(2))-1=(v^(2))/(1-v^(2))=>k=(va_(0))/(sqrt(1-v^(2)))]:}\begin{align*} & g(\dot{\gamma}, \dot{\gamma})=1=\dot{t}^{2}-a_{0}^{2} \dot{\varphi}^{2}=\frac{1}{1-v^{2}}-a_{0}^{2} \dot{\varphi}^{2}=\frac{1}{1-v^{2}}-\frac{k}{a_{0}^{2}} \\ & \Rightarrow \quad \frac{k^{2}}{a_{0}^{2}}=\frac{1}{1-v^{2}}-1=\frac{v^{2}}{1-v^{2}} \Rightarrow k=\frac{v a_{0}}{\sqrt{1-v^{2}}} \tag{3.2049} \end{align*}g(γ˙,γ˙)=1=t˙2a02φ˙2=11v2a02φ˙2=11v2ka02(3.2049)k2a02=11v21=v21v2k=va01v2
From the 4 -velocity having constant magnitude equals to 1 , we deduce
(3.2050) 1 = t ˙ 2 a 2 φ ˙ 2 = t ˙ 2 k 2 a 2 t ˙ = 1 + k 2 a 2 (3.2050) 1 = t ˙ 2 a 2 φ ˙ 2 = t ˙ 2 k 2 a 2 t ˙ = 1 + k 2 a 2 {:(3.2050)1=t^(˙)^(2)-a^(2)varphi^(˙)^(2)=t^(˙)^(2)-(k^(2))/(a^(2))=>t^(˙)=sqrt(1+(k^(2))/(a^(2))):}\begin{equation*} 1=\dot{t}^{2}-a^{2} \dot{\varphi}^{2}=\dot{t}^{2}-\frac{k^{2}}{a^{2}} \Rightarrow \dot{t}=\sqrt{1+\frac{k^{2}}{a^{2}}} \tag{3.2050} \end{equation*}(3.2050)1=t˙2a2φ˙2=t˙2k2a2t˙=1+k2a2
Dividing this by φ ˙ = k / a 2 φ ˙ = k / a 2 varphi^(˙)=k//a^(2)\dot{\varphi}=k / a^{2}φ˙=k/a2 leads to the separable differential equation
(3.2051) d t d φ = i ˙ φ ˙ = a 2 1 + k 2 / a 2 k = a k a 2 + k 2 (3.2051) d t d φ = i ˙ φ ˙ = a 2 1 + k 2 / a 2 k = a k a 2 + k 2 {:(3.2051)(dt)/(d varphi)=((i^(˙)))/((varphi^(˙)))=(a^(2)sqrt(1+k^(2)//a^(2)))/(k)=(a)/(k)sqrt(a^(2)+k^(2)):}\begin{equation*} \frac{d t}{d \varphi}=\frac{\dot{i}}{\dot{\varphi}}=\frac{a^{2} \sqrt{1+k^{2} / a^{2}}}{k}=\frac{a}{k} \sqrt{a^{2}+k^{2}} \tag{3.2051} \end{equation*}(3.2051)dtdφ=i˙φ˙=a21+k2/a2k=aka2+k2
In order for the object to complete an entire lap around the universe in finite time t 1 t 1 t_(1)t_{1}t1, we therefore require that
(3.2052) 2 π = 0 2 π d φ = t 0 t 1 k a k 2 + a 2 d t (3.2052) 2 π = 0 2 π d φ = t 0 t 1 k a k 2 + a 2 d t {:(3.2052)2pi=int_(0)^(2pi)d varphi=int_(t_(0))^(t_(1))(k)/(asqrt(k^(2)+a^(2)))dt:}\begin{equation*} 2 \pi=\int_{0}^{2 \pi} d \varphi=\int_{t_{0}}^{t_{1}} \frac{k}{a \sqrt{k^{2}+a^{2}}} d t \tag{3.2052} \end{equation*}(3.2052)2π=02πdφ=t0t1kak2+a2dt
for some finite t 1 t 1 t_(1)t_{1}t1. In other words, we have
(3.2053) t 0 k a k 2 + a 2 d t > 2 π t 0 v a 0 a ( t ) d t ( 1 v 2 ) a ( t ) 2 + a 0 2 v 2 > 2 π (3.2053) t 0 k a k 2 + a 2 d t > 2 π t 0 v a 0 a ( t ) d t 1 v 2 a ( t ) 2 + a 0 2 v 2 > 2 π {:(3.2053)int_(t_(0))^(oo)(k)/(asqrt(k^(2)+a^(2)))dt > 2pi=>int_(t_(0))^(oo)(va_(0))/(a(t))(dt)/(sqrt((1-v^(2))a(t)^(2)+a_(0)^(2)v^(2))) > 2pi:}\begin{equation*} \int_{t_{0}}^{\infty} \frac{k}{a \sqrt{k^{2}+a^{2}}} d t>2 \pi \Rightarrow \int_{t_{0}}^{\infty} \frac{v a_{0}}{a(t)} \frac{d t}{\sqrt{\left(1-v^{2}\right) a(t)^{2}+a_{0}^{2} v^{2}}}>2 \pi \tag{3.2053} \end{equation*}(3.2053)t0kak2+a2dt>2πt0va0a(t)dt(1v2)a(t)2+a02v2>2π
b) We know that g ( γ ˙ , φ ) = a 2 φ ˙ = k g γ ˙ , φ = a 2 φ ˙ = k g((gamma^(˙)),del_(varphi))=-a^(2)varphi^(˙)=-kg\left(\dot{\gamma}, \partial_{\varphi}\right)=-a^{2} \dot{\varphi}=-kg(γ˙,φ)=a2φ˙=k is constant and that t ˙ = 1 + k 2 / a 2 t ˙ = 1 + k 2 / a 2 t^(˙)=sqrt(1+k^(2)//a^(2))\dot{t}=\sqrt{1+k^{2} / a^{2}}t˙=1+k2/a2. The relative velocity v ( t ) v ( t ) v(t)v(t)v(t) at time t t ttt is therefore given by
(3.2054) γ = g ( t , γ ˙ ) = 1 1 v ( t ) 2 (3.2054) γ = g t , γ ˙ = 1 1 v ( t ) 2 {:(3.2054)gamma=g(del_(t),(gamma^(˙)))=(1)/(sqrt(1-v(t)^(2))):}\begin{equation*} \gamma=g\left(\partial_{t}, \dot{\gamma}\right)=\frac{1}{\sqrt{1-v(t)^{2}}} \tag{3.2054} \end{equation*}(3.2054)γ=g(t,γ˙)=11v(t)2
We find that
(3.2055) γ = g ( t , γ ˙ ) = t ˙ = 1 + k 2 a ( t ) 2 = 1 + a 0 2 a ( t ) 2 v 2 1 v 2 (3.2055) γ = g t , γ ˙ = t ˙ = 1 + k 2 a ( t ) 2 = 1 + a 0 2 a ( t ) 2 v 2 1 v 2 {:(3.2055)gamma=g(del_(t),(gamma^(˙)))=t^(˙)=sqrt(1+(k^(2))/(a(t)^(2)))=sqrt(1+(a_(0)^(2))/(a(t)^(2))(v^(2))/(1-v^(2))):}\begin{equation*} \gamma=g\left(\partial_{t}, \dot{\gamma}\right)=\dot{t}=\sqrt{1+\frac{k^{2}}{a(t)^{2}}}=\sqrt{1+\frac{a_{0}^{2}}{a(t)^{2}} \frac{v^{2}}{1-v^{2}}} \tag{3.2055} \end{equation*}(3.2055)γ=g(t,γ˙)=t˙=1+k2a(t)2=1+a02a(t)2v21v2
Solving for v ( t ) v ( t ) v(t)v(t)v(t), we obtain
(3.2056) v ( t ) = a 0 v a ( t ) 2 + v 2 [ a 0 2 a ( t ) 2 ] (3.2056) v ( t ) = a 0 v a ( t ) 2 + v 2 a 0 2 a ( t ) 2 {:(3.2056)v(t)=(a_(0)v)/(sqrt(a(t)^(2)+v^(2)[a_(0)^(2)-a(t)^(2)])):}\begin{equation*} v(t)=\frac{a_{0} v}{\sqrt{a(t)^{2}+v^{2}\left[a_{0}^{2}-a(t)^{2}\right]}} \tag{3.2056} \end{equation*}(3.2056)v(t)=a0va(t)2+v2[a02a(t)2]
Note that v ( t 0 ) = v v t 0 = v v(t_(0))=vv\left(t_{0}\right)=vv(t0)=v and that v ( t ) 0 v ( t ) 0 v(t)rarr0v(t) \rightarrow 0v(t)0 as a ( t ) / a 0 a ( t ) / a 0 a(t)//a_(0)rarr ooa(t) / a_{0} \rightarrow \inftya(t)/a0.

2.147

For a flat universe containing matter and radiation only, the following must be satisfied
(3.2057) 1 = Ω m + Ω r = Ω m + x Ω m = 1 x (3.2057) 1 = Ω m + Ω r = Ω m + x Ω m = 1 x {:(3.2057)1=Omega_(m)+Omega_(r)=Omega_(m)+x quad=>quadOmega_(m)=1-x:}\begin{equation*} 1=\Omega_{m}+\Omega_{r}=\Omega_{m}+x \quad \Rightarrow \quad \Omega_{m}=1-x \tag{3.2057} \end{equation*}(3.2057)1=Ωm+Ωr=Ωm+xΩm=1x
The time evolution of the scale factor a ( t ) a ( t ) a(t)a(t)a(t) can be found through the relationship
(3.2058) a ˙ = 8 π G 3 ρ 0 , m a 1 + ρ 0 , r a 2 = H 0 Ω m a 1 + Ω r a 2 , (3.2058) a ˙ = 8 π G 3 ρ 0 , m a 1 + ρ 0 , r a 2 = H 0 Ω m a 1 + Ω r a 2 , {:(3.2058)a^(˙)=sqrt((8pi G)/(3))sqrt(rho_(0,m)a^(-1)+rho_(0,r)a^(-2))=H_(0)sqrt(Omega_(m)a^(-1)+Omega_(r)a^(-2))",":}\begin{equation*} \dot{a}=\sqrt{\frac{8 \pi G}{3}} \sqrt{\rho_{0, m} a^{-1}+\rho_{0, r} a^{-2}}=H_{0} \sqrt{\Omega_{m} a^{-1}+\Omega_{r} a^{-2}}, \tag{3.2058} \end{equation*}(3.2058)a˙=8πG3ρ0,ma1+ρ0,ra2=H0Ωma1+Ωra2,
where it has been assumed that a 0 = 1 a 0 = 1 a_(0)=1a_{0}=1a0=1 and used that 8 π G 3 = H 0 2 ρ 0 , c 8 π G 3 = H 0 2 ρ 0 , c sqrt((8pi G)/(3))=(H_(0)^(2))/(rho_(0,c))\sqrt{\frac{8 \pi G}{3}}=\frac{H_{0}^{2}}{\rho_{0, c}}8πG3=H02ρ0,c. This is a separable differential equation and we find that
(3.2059) t 0 t 1 = t 1 t 0 d t = 1 H 0 a 1 a 0 d a Ω m a 1 + Ω r a 2 = 1 H 0 a 1 1 a d a ( 1 x ) a + x (3.2059) t 0 t 1 = t 1 t 0 d t = 1 H 0 a 1 a 0 d a Ω m a 1 + Ω r a 2 = 1 H 0 a 1 1 a d a ( 1 x ) a + x {:(3.2059)t_(0)-t_(1)=int_(t_(1))^(t_(0))dt=(1)/(H_(0))int_(a_(1))^(a_(0))(da)/(sqrt(Omega_(m)a^(-1)+Omega_(r)a^(-2)))=(1)/(H_(0))int_(a_(1))^(1)(ada)/(sqrt((1-x)a+x)):}\begin{equation*} t_{0}-t_{1}=\int_{t_{1}}^{t_{0}} d t=\frac{1}{H_{0}} \int_{a_{1}}^{a_{0}} \frac{d a}{\sqrt{\Omega_{m} a^{-1}+\Omega_{r} a^{-2}}}=\frac{1}{H_{0}} \int_{a_{1}}^{1} \frac{a d a}{\sqrt{(1-x) a+x}} \tag{3.2059} \end{equation*}(3.2059)t0t1=t1t0dt=1H0a1a0daΩma1+Ωra2=1H0a11ada(1x)a+x
since we work under the assumption that a 0 = 1 a 0 = 1 a_(0)=1a_{0}=1a0=1. Performing this integral, e.g., by partial integration, leads to
(3.2060) t 0 t 1 = 1 H 0 2 3 [ ( 1 x ) a + x ( 1 x ) 2 [ ( 1 x ) a 2 x ] ] a 1 1 (3.2060) t 0 t 1 = 1 H 0 2 3 ( 1 x ) a + x ( 1 x ) 2 [ ( 1 x ) a 2 x ] a 1 1 {:(3.2060)t_(0)-t_(1)=(1)/(H_(0))(2)/(3)[(sqrt((1-x)a+x))/((1-x)^(2))[(1-x)a-2x]]_(a_(1))^(1):}\begin{equation*} t_{0}-t_{1}=\frac{1}{H_{0}} \frac{2}{3}\left[\frac{\sqrt{(1-x) a+x}}{(1-x)^{2}}[(1-x) a-2 x]\right]_{a_{1}}^{1} \tag{3.2060} \end{equation*}(3.2060)t0t1=1H023[(1x)a+x(1x)2[(1x)a2x]]a11
The energy density ratio between matter and radiation scales as ρ m / ρ r a ( t ) ρ m / ρ r a ( t ) rho_(m)//rho_(r)∼a(t)\rho_{m} / \rho_{r} \sim a(t)ρm/ρra(t) due to ρ m a 3 ρ m a 3 rho_(m)propa^(-3)\rho_{m} \propto a^{-3}ρma3 and ρ r a 4 ρ r a 4 rho_(r)propa^(-4)\rho_{r} \propto a^{-4}ρra4. Since
(3.2061) ρ 0 , m ρ 0 , r = 1 x x 1 x (3.2061) ρ 0 , m ρ 0 , r = 1 x x 1 x {:(3.2061)(rho_(0,m))/(rho_(0,r))=(1-x)/(x)≃(1)/(x):}\begin{equation*} \frac{\rho_{0, m}}{\rho_{0, r}}=\frac{1-x}{x} \simeq \frac{1}{x} \tag{3.2061} \end{equation*}(3.2061)ρ0,mρ0,r=1xx1x
we find that ρ m ρ r ρ m ρ r rho_(m)≃rho_(r)\rho_{m} \simeq \rho_{r}ρmρr when
(3.2062) ρ m ρ r = ρ 0 , m ρ 0 , r a ( t 1 ) a 0 = a ( t 1 ) x a ( t 1 ) x (3.2062) ρ m ρ r = ρ 0 , m ρ 0 , r a t 1 a 0 = a t 1 x a t 1 x {:(3.2062)(rho_(m))/(rho_(r))=(rho_(0,m))/(rho_(0,r))(a(t_(1)))/(a_(0))=(a(t_(1)))/(x)=>a(t_(1))≃x:}\begin{equation*} \frac{\rho_{m}}{\rho_{r}}=\frac{\rho_{0, m}}{\rho_{0, r}} \frac{a\left(t_{1}\right)}{a_{0}}=\frac{a\left(t_{1}\right)}{x} \Rightarrow a\left(t_{1}\right) \simeq x \tag{3.2062} \end{equation*}(3.2062)ρmρr=ρ0,mρ0,ra(t1)a0=a(t1)xa(t1)x
It follows that
t 0 t 1 = 2 3 H 0 { ( 1 x ) + x ( 1 x ) 2 [ ( 1 x ) 2 x ] ( 1 x ) x + x ( 1 x ) 2 [ ( 1 x ) x 2 x ] } (3.2063) 2 3 H 0 [ ( 1 + 2 x ) ( 1 3 x ) ] 2 3 H 0 ( 1 x ) t 0 t 1 = 2 3 H 0 ( 1 x ) + x ( 1 x ) 2 [ ( 1 x ) 2 x ] ( 1 x ) x + x ( 1 x ) 2 [ ( 1 x ) x 2 x ] (3.2063) 2 3 H 0 [ ( 1 + 2 x ) ( 1 3 x ) ] 2 3 H 0 ( 1 x ) {:[t_(0)-t_(1)=(2)/(3H_(0)){(sqrt((1-x)+x))/((1-x)^(2))[(1-x)-2x]-(sqrt((1-x)x+x))/((1-x)^(2))*[(1-x)x-2x]}],[(3.2063)≃(2)/(3H_(0))[(1+2x)(1-3x)]≃(2)/(3H_(0))(1-x)]:}\begin{align*} t_{0}-t_{1} & =\frac{2}{3 H_{0}}\left\{\frac{\sqrt{(1-x)+x}}{(1-x)^{2}}[(1-x)-2 x]-\frac{\sqrt{(1-x) x+x}}{(1-x)^{2}} \cdot[(1-x) x-2 x]\right\} \\ & \simeq \frac{2}{3 H_{0}}[(1+2 x)(1-3 x)] \simeq \frac{2}{3 H_{0}}(1-x) \tag{3.2063} \end{align*}t0t1=23H0{(1x)+x(1x)2[(1x)2x](1x)x+x(1x)2[(1x)x2x]}(3.2063)23H0[(1+2x)(13x)]23H0(1x)
The time since matter-radiation equality is therefore just slightly less than the age 2 3 H 0 2 3 H 0 (2)/(3H_(0))\frac{2}{3 H_{0}}23H0 of a fully matter-dominated universe.
2.148
The curvature parameter is given by
(3.2064) | Ω K | = | 1 H 2 a 2 | = | 1 ( a ˙ a ) 2 a 2 | = 1 a ˙ 2 (3.2064) Ω K = 1 H 2 a 2 = 1 a ˙ a 2 a 2 = 1 a ˙ 2 {:(3.2064)|Omega_(K)|=|-(1)/(H^(2)a^(2))|=|-(1)/((((a^(˙)))/(a))^(2)a^(2))|=(1)/(a^(˙)^(2)):}\begin{equation*} \left|\Omega_{K}\right|=\left|-\frac{1}{H^{2} a^{2}}\right|=\left|-\frac{1}{\left(\frac{\dot{a}}{a}\right)^{2} a^{2}}\right|=\frac{1}{\dot{a}^{2}} \tag{3.2064} \end{equation*}(3.2064)|ΩK|=|1H2a2|=|1(a˙a)2a2|=1a˙2
The time derivative of | Ω K | Ω K |Omega_(K)|\left|\Omega_{K}\right||ΩK| is therefore
(3.2065) d | Ω K | d t = d ( 1 / a ˙ 2 ) d t = 2 a ˙ 3 a ¨ (3.2065) d Ω K d t = d 1 / a ˙ 2 d t = 2 a ˙ 3 a ¨ {:(3.2065)(d|Omega_(K)|)/(dt)=(d(1//a^(˙)^(2)))/(dt)=-(2)/(a^(˙)^(3))a^(¨):}\begin{equation*} \frac{d\left|\Omega_{K}\right|}{d t}=\frac{d\left(1 / \dot{a}^{2}\right)}{d t}=-\frac{2}{\dot{a}^{3}} \ddot{a} \tag{3.2065} \end{equation*}(3.2065)d|ΩK|dt=d(1/a˙2)dt=2a˙3a¨
For an expanding universe, a ˙ > 0 a ˙ > 0 a^(˙) > 0\dot{a}>0a˙>0, which implies that a ˙ 3 > 0 a ˙ 3 > 0 a^(˙)^(3) > 0\dot{a}^{3}>0a˙3>0, and thus, the requirement that | Ω K | Ω K |Omega_(K)|\left|\Omega_{K}\right||ΩK| decreases with time is given by
(3.2066) d | Ω K | d t = 2 a ˙ 3 a ¨ < 0 a ¨ > 0 (3.2066) d Ω K d t = 2 a ˙ 3 a ¨ < 0 a ¨ > 0 {:(3.2066)(d|Omega_(K)|)/(dt)=-(2)/(a^(˙)^(3))a^(¨) < 0quad=>quada^(¨) > 0:}\begin{equation*} \frac{d\left|\Omega_{K}\right|}{d t}=-\frac{2}{\dot{a}^{3}} \ddot{a}<0 \quad \Rightarrow \quad \ddot{a}>0 \tag{3.2066} \end{equation*}(3.2066)d|ΩK|dt=2a˙3a¨<0a¨>0
That | Ω K | Ω K |Omega_(K)|\left|\Omega_{K}\right||ΩK| decreases with time is therefore equivalent to the scale factor growth accelerating. The Friedmann acceleration equation is
(3.2067) a ¨ = 4 π G 3 a ( ρ + 3 p ) > 0 ρ + 3 p < 0 p ρ < 1 3 . (3.2067) a ¨ = 4 π G 3 a ( ρ + 3 p ) > 0 ρ + 3 p < 0 p ρ < 1 3 . {:(3.2067)a^(¨)=-(4pi G)/(3)a(rho+3p) > 0=>rho+3p < 0quad=>quad(p)/( rho) < -(1)/(3).:}\begin{equation*} \ddot{a}=-\frac{4 \pi G}{3} a(\rho+3 p)>0 \Rightarrow \rho+3 p<0 \quad \Rightarrow \quad \frac{p}{\rho}<-\frac{1}{3} . \tag{3.2067} \end{equation*}(3.2067)a¨=4πG3a(ρ+3p)>0ρ+3p<0pρ<13.
The condition on w w www for | Ω K | Ω K |Omega_(K)|\left|\Omega_{K}\right||ΩK| to decrease with time is therefore w < 1 3 w < 1 3 w < -(1)/(3)w<-\frac{1}{3}w<13. In particular, note that this is the case for a cosmological constant, where w = 1 w = 1 w=-1w=-1w=1.
2.149
a) From the evolution equations, we find that
(3.2068) a ˙ = 8 π G ρ 0 , c 3 a 1 + 3 w 2 = H 0 a k (3.2068) a ˙ = 8 π G ρ 0 , c 3 a 1 + 3 w 2 = H 0 a k {:(3.2068)a^(˙)=sqrt((8pi Grho_(0,c))/(3))a^(-(1+3w)/(2))=H_(0)a^(-k):}\begin{equation*} \dot{a}=\sqrt{\frac{8 \pi G \rho_{0, c}}{3}} a^{-\frac{1+3 w}{2}}=H_{0} a^{-k} \tag{3.2068} \end{equation*}(3.2068)a˙=8πGρ0,c3a1+3w2=H0ak
where k = 1 2 + 3 w 2 k = 1 2 + 3 w 2 k=(1)/(2)+(3w)/(2)k=\frac{1}{2}+\frac{3 w}{2}k=12+3w2. It follows that a ˙ a k = H 0 a ˙ a k = H 0 a^(˙)a^(k)=H_(0)\dot{a} a^{k}=H_{0}a˙ak=H0, and therefore,
H 0 ( t t 0 ) = t 0 t H 0 d t = a 0 a ( t ) a k d a = 1 a ( t ) a k d a (3.2069) = { a ( t ) k + 1 1 k + 1 , k 1 w 1 ln a ( t ) , k = 1 w = 1 . H 0 t t 0 = t 0 t H 0 d t = a 0 a ( t ) a k d a = 1 a ( t ) a k d a (3.2069) = a ( t ) k + 1 1 k + 1 , k 1 w 1 ln a ( t ) , k = 1 w = 1 . {:[H_(0)(t-t_(0))=int_(t_(0))^(t)H_(0)dt=int_(a_(0))^(a(t))a^(k)da=int_(1)^(a(t))a^(k)da],[(3.2069)={[(a(t)^(k+1)-1)/(k+1)","quad k!=-1harr w!=-1],[ln a(t)",",k=-1harr w=-1]:}]:}.\begin{align*} H_{0}\left(t-t_{0}\right)=\int_{t_{0}}^{t} H_{0} d t=\int_{a_{0}}^{a(t)} a^{k} d a & =\int_{1}^{a(t)} a^{k} d a \\ & = \begin{cases}\frac{a(t)^{k+1}-1}{k+1}, \quad k \neq-1 \leftrightarrow w \neq-1 \\ \ln a(t), & k=-1 \leftrightarrow w=-1\end{cases} \tag{3.2069} \end{align*} .H0(tt0)=t0tH0dt=a0a(t)akda=1a(t)akda(3.2069)={a(t)k+11k+1,k1w1lna(t),k=1w=1.
Solving for a ( t ) a ( t ) a(t)a(t)a(t) leads to
(3.2070) a ( t ) = { [ ( k + 1 ) H 0 ( t t 0 ) + 1 ] 2 ( 1 + w ) 2 , w 1 exp H 0 ( t t 0 ) , w = 1 (3.2070) a ( t ) = ( k + 1 ) H 0 t t 0 + 1 2 ( 1 + w ) 2 , w 1 exp H 0 t t 0 , w = 1 {:(3.2070)a(t)={[[(k+1)H_(0)(t-t_(0))+1]^((2(1+w))/(2))","quad w!=-1],[exp H_(0)(t-t_(0))","quad w=1]:}:}a(t)=\left\{\begin{array}{l} {\left[(k+1) H_{0}\left(t-t_{0}\right)+1\right]^{\frac{2(1+w)}{2}}, \quad w \neq-1} \tag{3.2070}\\ \exp H_{0}\left(t-t_{0}\right), \quad w=1 \end{array}\right.(3.2070)a(t)={[(k+1)H0(tt0)+1]2(1+w)2,w1expH0(tt0),w=1
b) For the Hubble parameter, we find that
(3.2071) H ( t ) = a ˙ a = H 0 a k + 1 = H 0 1 + 3 2 ( 1 + w ) H 0 ( t t 0 ) , w 1 . (3.2071) H ( t ) = a ˙ a = H 0 a k + 1 = H 0 1 + 3 2 ( 1 + w ) H 0 t t 0 , w 1 . {:(3.2071)H(t)=((a^(˙)))/(a)=H_(0)a^(k+1)=(H_(0))/(1+(3)/(2)(1+w)H_(0)(t-t_(0)))","quad w!=-1.:}\begin{equation*} H(t)=\frac{\dot{a}}{a}=H_{0} a^{k+1}=\frac{H_{0}}{1+\frac{3}{2}(1+w) H_{0}\left(t-t_{0}\right)}, \quad w \neq-1 . \tag{3.2071} \end{equation*}(3.2071)H(t)=a˙a=H0ak+1=H01+32(1+w)H0(tt0),w1.
For w = 1 w = 1 w=-1w=-1w=1, we find that H ( t ) = H 0 H ( t ) = H 0 H(t)=H_(0)H(t)=H_{0}H(t)=H0, which is also equal to the above expression with w = 1 w = 1 w=-1w=-1w=1 inserted. Thus, generally
(3.2072) H ( t ) = H 0 1 + 3 2 ( 1 + w ) H 0 ( t t 0 ) (3.2072) H ( t ) = H 0 1 + 3 2 ( 1 + w ) H 0 t t 0 {:(3.2072)H(t)=(H_(0))/(1+(3)/(2)(1+w)H_(0)(t-t_(0))):}\begin{equation*} H(t)=\frac{H_{0}}{1+\frac{3}{2}(1+w) H_{0}\left(t-t_{0}\right)} \tag{3.2072} \end{equation*}(3.2072)H(t)=H01+32(1+w)H0(tt0)
Note that H ( t ) H ( t ) H(t)H(t)H(t) decreases with time whenever w > 1 w > 1 w > -1w>-1w>1.

2.150

a) The matter part of the action is given by
(3.2073) S m = L g ¯ | d 4 x = [ g μ v ( μ ϕ ) ( ν ϕ ) V ( ϕ ) ] g ¯ | d 4 x (3.2073) S m = L g ¯ d 4 x = g μ v μ ϕ ν ϕ V ( ϕ ) g ¯ d 4 x {:(3.2073)S_(m)=intLsqrt(∣ bar(g))|d^(4)x=int[g^(mu v)(del_(mu)phi)(del_(nu)phi)-V(phi)]sqrt(∣ bar(g))|d^(4)x:}\begin{equation*} \mathcal{S}_{m}=\int \mathcal{L} \sqrt{\mid \bar{g}}\left|d^{4} x=\int\left[g^{\mu v}\left(\partial_{\mu} \phi\right)\left(\partial_{\nu} \phi\right)-V(\phi)\right] \sqrt{\mid \bar{g}}\right| d^{4} x \tag{3.2073} \end{equation*}(3.2073)Sm=Lg¯|d4x=[gμv(μϕ)(νϕ)V(ϕ)]g¯|d4x
Taking the variation of this leads to
δ S m = [ 1 2 ( μ ϕ ) ( ν ϕ ) δ g μ ν + g μ ν ( μ ϕ ) ( ν δ ϕ ) V ( ϕ ) δ ϕ ] g ¯ | d 4 x (3.2074) + L δ | g ¯ | d 4 x δ S m = 1 2 μ ϕ ν ϕ δ g μ ν + g μ ν μ ϕ ν δ ϕ V ( ϕ ) δ ϕ g ¯ d 4 x (3.2074) + L δ | g ¯ | d 4 x {:[deltaS_(m)={: int[(1)/(2)(del_(mu)phi)(del_(nu)phi)deltag^(mu nu)+g^(mu nu)(del_(mu)phi)(del_(nu)delta phi)-V^(')(phi)delta phi]sqrt(∣ bar(g))|d^(4)x],[(3.2074)+intLdeltasqrt(| bar(g)|)d^(4)x]:}\begin{align*} \delta \mathcal{S}_{m}= & \left.\int\left[\frac{1}{2}\left(\partial_{\mu} \phi\right)\left(\partial_{\nu} \phi\right) \delta g^{\mu \nu}+g^{\mu \nu}\left(\partial_{\mu} \phi\right)\left(\partial_{\nu} \delta \phi\right)-V^{\prime}(\phi) \delta \phi\right] \sqrt{\mid \bar{g}} \right\rvert\, d^{4} x \\ & +\int \mathcal{L} \delta \sqrt{|\bar{g}|} d^{4} x \tag{3.2074} \end{align*}δSm=[12(μϕ)(νϕ)δgμν+gμν(μϕ)(νδϕ)V(ϕ)δϕ]g¯|d4x(3.2074)+Lδ|g¯|d4x
The equation of motion for the scalar field ϕ ϕ phi\phiϕ itself is found by varying the action with respect to ϕ ϕ phi\phiϕ only. From the variation of the action derived in a), we then find
(3.2075) δ S m = [ g μ ν ( μ ϕ ) ( ν δ ϕ ) V ( ϕ ) δ ϕ ] g ¯ d 4 x = 0 (3.2075) δ S m = g μ ν μ ϕ ν δ ϕ V ( ϕ ) δ ϕ g ¯ d 4 x = 0 {:(3.2075)deltaS_(m)=int[g^(mu nu)(del_(mu)phi)(del_(nu)delta phi)-V^(')(phi)delta phi]sqrt(∣ bar(g))∣d^(4)x=0:}\begin{equation*} \delta \mathcal{S}_{m}=\int\left[g^{\mu \nu}\left(\partial_{\mu} \phi\right)\left(\partial_{\nu} \delta \phi\right)-V^{\prime}(\phi) \delta \phi\right] \sqrt{\mid \bar{g}} \mid d^{4} x=0 \tag{3.2075} \end{equation*}(3.2075)δSm=[gμν(μϕ)(νδϕ)V(ϕ)δϕ]g¯d4x=0
Partial integration of the first term leads to
(3.2076) δ S m = δ ϕ [ v { g μ ν | g ¯ | μ ϕ } + V ( ϕ ) | g ¯ | ] d 4 x = 0 (3.2076) δ S m = δ ϕ v g μ ν | g ¯ | μ ϕ + V ( ϕ ) | g ¯ | d 4 x = 0 {:(3.2076)deltaS_(m)=-int delta phi[del_(v){g^(mu nu)sqrt(| bar(g)|)del_(mu)phi}+V^(')(phi)sqrt(| bar(g)|)]d^(4)x=0:}\begin{equation*} \delta \mathcal{S}_{m}=-\int \delta \phi\left[\partial_{v}\left\{g^{\mu \nu} \sqrt{|\bar{g}|} \partial_{\mu} \phi\right\}+V^{\prime}(\phi) \sqrt{|\bar{g}|}\right] d^{4} x=0 \tag{3.2076} \end{equation*}(3.2076)δSm=δϕ[v{gμν|g¯|μϕ}+V(ϕ)|g¯|]d4x=0
Since this should hold regardless of the variation δ ϕ δ ϕ delta phi\delta \phiδϕ, we conclude that
(3.2077) 1 | g ¯ | ν ( g μ v g ¯ μ ϕ ) = g μ v μ ν ϕ = V ( ϕ ) (3.2077) 1 | g ¯ | ν g μ v g ¯ μ ϕ = g μ v μ ν ϕ = V ( ϕ ) {:(3.2077)(1)/(sqrt(| bar(g)|))del_(nu)(g^(mu v)sqrt(∣ bar(g))∣del_(mu)phi)=g^(mu v)grad_(mu)grad_(nu)phi=-V^(')(phi):}\begin{equation*} \frac{1}{\sqrt{|\bar{g}|}} \partial_{\nu}\left(g^{\mu v} \sqrt{\mid \bar{g}} \mid \partial_{\mu} \phi\right)=g^{\mu v} \nabla_{\mu} \nabla_{\nu} \phi=-V^{\prime}(\phi) \tag{3.2077} \end{equation*}(3.2077)1|g¯|ν(gμvg¯μϕ)=gμvμνϕ=V(ϕ)
This is the equation of motion for ϕ ϕ phi\phiϕ.
b) With i ϕ = 0 i ϕ = 0 del_(i)phi=0\partial_{i} \phi=0iϕ=0 and t ϕ = ϕ ˙ t ϕ = ϕ ˙ del_(t)phi=phi^(˙)\partial_{t} \phi=\dot{\phi}tϕ=ϕ˙ and using the result of Problem 2.60, we find that
(3.2078) T μ ν = U μ U ν ϕ ˙ 2 g μ ν [ 1 2 ϕ ˙ 2 V ( ϕ ) ] , (3.2078) T μ ν = U μ U ν ϕ ˙ 2 g μ ν 1 2 ϕ ˙ 2 V ( ϕ ) , {:(3.2078)T_(mu nu)=U_(mu)U_(nu)phi^(˙)^(2)-g_(mu nu)[(1)/(2)phi^(˙)^(2)-V(phi)]",":}\begin{equation*} T_{\mu \nu}=U_{\mu} U_{\nu} \dot{\phi}^{2}-g_{\mu \nu}\left[\frac{1}{2} \dot{\phi}^{2}-V(\phi)\right], \tag{3.2078} \end{equation*}(3.2078)Tμν=UμUνϕ˙2gμν[12ϕ˙2V(ϕ)],
where U μ = δ μ 0 U μ = δ μ 0 U_(mu)=delta_(mu)^(0)U_{\mu}=\delta_{\mu}^{0}Uμ=δμ0. We note that g μ ν U μ U ν = 1 g μ ν U μ U ν = 1 g^(mu nu)U_(mu)U_(nu)=1g^{\mu \nu} U_{\mu} U_{\nu}=1gμνUμUν=1 and U U UUU is therefore a 4-velocity field. Comparing this to the stress-energy tensor of an ideal fluid T μ ν = U μ U ν ( ρ 0 + p ) T μ ν = U μ U ν ρ 0 + p T_(mu nu)=U_(mu)U_(nu)(rho_(0)+p)-T_{\mu \nu}=U_{\mu} U_{\nu}\left(\rho_{0}+p\right)-Tμν=UμUν(ρ0+p) g μ ν p g μ ν p g_(mu nu)pg_{\mu \nu} pgμνp, we obtain
(3.2079) p = 1 2 ϕ ˙ 2 V ( ϕ ) and ρ 0 = 1 2 ϕ ˙ 2 + V ( ϕ ) (3.2079) p = 1 2 ϕ ˙ 2 V ( ϕ )  and  ρ 0 = 1 2 ϕ ˙ 2 + V ( ϕ ) {:(3.2079)p=(1)/(2)*phi^(˙)^(2)-V(phi)quad" and "quadrho_(0)=(1)/(2)*phi^(˙)^(2)+V(phi):}\begin{equation*} p=\frac{1}{2} \cdot \dot{\phi}^{2}-V(\phi) \quad \text { and } \quad \rho_{0}=\frac{1}{2} \cdot \dot{\phi}^{2}+V(\phi) \tag{3.2079} \end{equation*}(3.2079)p=12ϕ˙2V(ϕ) and ρ0=12ϕ˙2+V(ϕ)
The equation-of-state parameter is therefore given by
(3.2080) w = p ρ 0 = ϕ ˙ 2 2 V ( ϕ ) ϕ ˙ 2 + 2 V ( ϕ ) (3.2080) w = p ρ 0 = ϕ ˙ 2 2 V ( ϕ ) ϕ ˙ 2 + 2 V ( ϕ ) {:(3.2080)w=(p)/(rho_(0))=(phi^(˙)^(2)-2V(phi))/(phi^(˙)^(2)+2V(phi)):}\begin{equation*} w=\frac{p}{\rho_{0}}=\frac{\dot{\phi}^{2}-2 V(\phi)}{\dot{\phi}^{2}+2 V(\phi)} \tag{3.2080} \end{equation*}(3.2080)w=pρ0=ϕ˙22V(ϕ)ϕ˙2+2V(ϕ)
Note that when ϕ ˙ 2 V ( ϕ ) ϕ ˙ 2 V ( ϕ ) phi^(˙)^(2)≪V(phi)\dot{\phi}^{2} \ll V(\phi)ϕ˙2V(ϕ), we find w 1 w 1 w≃-1w \simeq-1w1. This is the equation-of-state parameter for a cosmological constant.